Chapter 06 - Integration Techniques

Integration by substitution (review only)

Videos

Review Videos

[Note: this section is non-examinable. It is included for comparison to IBP.]

• Integration by Substitution 1: $\int \frac{-x}{(x+1)-\sqrt{x+1}}dx$
• Integration by Substitution 2: $\int \frac{x^5}{(1-x^3{)}^3}dx$
• Integration by Substitution 3: ${\int }_0^1{x}^2(1+x{)}^{4}dx$
• Integration by Substitution 4: $\int \frac{2x+3}{\sqrt{2x+1}}dx$
• Integration by Substitution 5: $\int \frac{\sin x}{{\cos }^{3}x}dx$
• Integration by Substitution: Definite integrals, various examples

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Theory

Theory 1

The method of $u$-substitution is applicable when the integrand is a product, with one factor a composite whose inner function’s derivative is the other factor.

Substitution

Suppose the integral has this format, for some functions $f$ and $u$:

$$\int f(u(x))\cdot u^{\prime }(x)dx$$

Then the rule says we may convert the integral into terms of $u$ considered as a variable, like this:

$$\int f(u(x))\cdot u^{\prime }(x)dx\gg \gg \int f(u)du$$

The technique of $u$-substitution comes from the chain rule for derivatives:

$$\frac{d}{dx}F(u(x))=f(u(x))\cdot u^{\prime }(x)$$

Here we let $F^{\prime }=f$. Thus $\int f(x)dx=F(x)+C$ for some $C$.

Now, if we integrate both sides of this equation, we find:

$$F(u(x))=\int f(u(x))\cdot u^{\prime }(x)dx$$

And of course $F\left(u\right)=\int f(u)du-C$.

Extra - Full explanation of $u$-substitution

(1) Chain rule for derivatives:

Let $F(x)$ be a function and $F^{\prime }=f$ its derivative. Let $u(x)$ be another function.

Using primes:

$$(F\circ u{)}^{\prime }=(F^{\prime }\circ u)\cdot u^{\prime }$$

Using differentials:

$$\frac{d}{dx}F(u(x))=f(u(x))\cdot u^{\prime }(x)$$

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(2) Integrate both sides:

$$\begin{array}{rl}\frac{d}{dx}F(u(x))=f(u(x))\cdot u^{\prime }(x)& \stackrel{\int }{\gg \gg }\int \frac{d}{dx}F(u(x))=\int f(u(x))\cdot u^{\prime }(x)\\ & \\ & \stackrel{\text{FTC}}{\gg \gg }F(u(x))=\int f(u(x))\cdot u^{\prime }(x)\end{array}$$

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(3) Introduce ‘variable’ $u$ from the $u$-format of the integral:

Treating $u$ as a variable, the definition of $F$ gives:

$$F(u)=\int f(u)du+C$$

Set the ‘variable’ $u$ to the ‘function’ $u$ output:

$$F(u){|}_{u=u(x)}=F(u(x))$$

Combine these:

$$\begin{array}{rl}F(u(x))& =F(u){|}_{u=u(x)}\\ & \\ & ={\int f(u)du|}_{u=u(x)}+C\end{array}$$

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(4) Substitute for $F(u(x))$:

$$\int f(u(x))\cdot u^{\prime }(x)dx=F(u(x))={\int f(u)du|}_{u=u(x)}+C$$

This is “$u$-substitution” in final form.

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Integration by parts

Videos

Review Videos

Videos:

• Integration by Parts 1: $\int e^{x}dx$ and $\int \ln xdx$
• Integration by Parts 2: $\int {\tan }^{-1}xdx$ and $\int x\sec xdx$
• Integration by Parts 3: Definite integrals
• Example: $\int e^{3x}\cos 4xdx$, two methods:
  • Double IBP
  • Fast Solution
• Integration by Parts 6: $\int {\sec }^{5}xdx$

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Theory

Theory 1

The method of integration by parts (abbreviated IBP) is applicable when the integrand is a product for which one factor is easily integrated while the other becomes simpler when differentiated.

Integration by parts

Suppose the integral has this format, for some functions $u$ and $v$:

$$\int u\cdot v^{\prime }dx$$

Then the rule says we may convert the integral like this:

$$\int u\cdot v^{\prime }dx\gg \gg u\cdot v-\int u^{\prime }\cdot vdx$$

This technique comes from the product rule for derivatives:

$$(u\cdot v{)}^{\prime }=u^{\prime }\cdot v+u\cdot v^{\prime }$$

Now, if we integrate both sides of this equation, we find:

$$u\cdot v=\int u^{\prime }\cdot vdx+\int u\cdot v^{\prime }dx$$

and the IBP rule follows by algebra.

Extra - Full explanation of integration by parts

(1) Product rule for derivatives:

$$(u\cdot v{)}^{\prime }=u^{\prime }\cdot v+u\cdot v^{\prime }$$

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(2) Integrate both sides:

$$\begin{array}{rl}(u\cdot v{)}^{\prime }=u^{\prime }\cdot v+u\cdot v^{\prime }& \gg \gg \int (u\cdot v{)}^{\prime }dx=\int u^{\prime }\cdot vdx+\int u\cdot v^{\prime }dx\\ & \\ & \gg \gg u\cdot v=\int u^{\prime }\cdot vdx+\int u\cdot v^{\prime }dx\\ & \\ & \gg \gg \int u\cdot v^{\prime }dx=u\cdot v-\int u^{\prime }\cdot v\end{array}\text{\hspace{1em}(FTC)(Rearrange)}$$

Definite IBP

Definite version of FTC:

$$\begin{array}{c}{\int }_a^b(u\cdot v{)}^{\prime }dx=u\cdot v{|}_a^b\\ \\ \gg \gg {\int }_a^{b}u\cdot v^{\prime }dx=u\cdot v{|}_a^b-{\int }_a^b{u}^{\prime }\cdot v\end{array}$$

Choosing factors well

IBP is symmetrical. How do we know which factor to choose for $u$ and which for $v$?

Here is a trick: the acronym “LIATE” spells out the order of choices – to the left for $u$ and to the right for $v$:

$$\begin{array}{c}\text{LIATE}:\\ \\ u\leftarrow \text{Logarithmic – Inverse\_trig – Algebraic – Trig – Exponential}\to v\end{array}$$

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Illustration

Example - A and T factors

A and T factors

Compute the integral: $\int x\cos xdx$

Solution

(1) Choose $u$ and $v^{\prime }$:

Set $u(x)=x$ because $x$ simplifies when differentiated.

(By the trick: $x$ is Algebraic, i.e. more “$u$”, and $\cos x$ is Trig, more “$v$”.)

Remaining factor must be $v^{\prime }=\cos x$.

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(2) Compute $u^{\prime }$ and $v$:

$$\begin{array}{rl}u=x& \gg \gg u^{\prime }=1\\ & \\ v^{\prime }=\cos x& \gg \gg v=\sin x\end{array}$$

Key chart:

$$\begin{array}{cc}u=x& v^{\prime }=\cos x\\ u^{\prime }=1& v=\sin x\end{array}\begin{array}{l}⟶\int u\cdot v^{\prime }\text{original}\\ ⟶\int u^{\prime }\cdot v\text{final}\end{array}$$

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(3) Evaluate IBP formula:

$$\begin{array}{c}\int u^{\prime }vdx=uv-\int u{v}^{\prime }dx\\ \\ \gg \gg \int x\cos xdx=x\sin x-\int 1\cdot \sin xdx\\ \\ \gg \gg x\sin x+\cos x+C\end{array}$$

Why IBP?

The rationale of IBP is that $\int 1\cdot \sin xdx$ is easier to compute than $\int x\cos xdx$.

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Exercise - Hidden A

Hidden A

Compute the integral:

$$\int \ln xdx$$

Solution

(1) Choose $u=\ln x$:

Because Log is farthest left in LIATE.

Therefore we must choose $v^{\prime }(x)=1$.

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(2) Compute $u^{\prime }$ and $v$:

We have $u^{\prime }=\frac{1}{x}$ and $v=x$. Obtain chart:

$$\begin{array}{cc}u=\ln x& v^{\prime }=1\\ u^{\prime }=1/x& v=x\end{array}\begin{array}{l}⟶\int u\cdot v^{\prime }\text{original}\\ ⟶\int u^{\prime }\cdot v\text{final}\end{array}$$

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(3) Evaluate IBP formula:

$$\begin{array}{c}\int u^{\prime }vdx=uv-\int u{v}^{\prime }dx\\ \\ \gg \gg \int \ln x\cdot 1dx=x\ln x-\int \frac{1}{x}\cdot xdx\end{array}$$

Perform new integration:

$$-\int \frac{1}{x}\cdot xdx\gg \gg -\int 1dx\gg \gg -x+C$$

Final answer is: $x\ln x-x+C$

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Trig power products

Videos

Review Videos

Videos, Math Dr. Bob:

• Trig power products: $\int {\cos }^{m}x{\sin }^{n}xdx$
• Trig differing frequencies: $\int \cos mx\sin nxdx$
• Trig tan and sec: $\int {\tan }^{m}x{\sec }^{n}xdx$
• Secant power: $\int {\sec }^{5}xdx$

Videos, Organic Chemistry Tutor:

• Trig power product techniques
• Trig substitution

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Theory

Theory 1

Review: trig identities

• ${\sin }^{2}x+{\cos }^{2}x=1$
• ${\sin }^{2}x=\frac{1}{2}(1-\cos 2x)$
• ${\cos }^{2}x=\frac{1}{2}(1+\cos 2x)$

Trig power product: $\sin /\cos$

A $\sin /\cos$ power product has this form:

$$\int {\cos }^{m}x\cdot {\sin }^{n}xdx$$

for some integers $m$ and $n$ (even negative!).

To compute these integrals, use a sequence of these techniques:

• Swap an even bunch.
• $u$-sub for power-one.
• Power-to-frequency conversion.

Memorize these three techniques!

Examples of trig power products:

• $\int \sin x\cdot {\cos }^{7}xdx$
• $\int {\sin }^{3}xdx$
• $\int {\sin }^{2}x\cdot {\cos }^{2}xdx$

Swap an even bunch

If either ${\cos }^{m}x$ or ${\sin }^{n}x$ is an odd power, use

$$\begin{array}{rl}{\sin }^{2}x\gg \gg & 1-{\cos }^{2}x\\ & \\ \text{OR}{\cos }^{2}x\gg \gg & 1-{\sin }^{2}x\end{array}$$

(maybe repeatedly) to convert an even bunch to the opposite trig type.

An even bunch is all but one from the odd power.

For example:

$$\begin{array}{rl}{\sin }^{5}x\cdot {\cos }^{8}x& \gg \gg \sin x({\sin }^{2}x{)}^2\cdot {\cos }^{8}x\\ & \\ & \gg \gg \sin x(1-{\cos }^{2}x{)}^2\cdot {\cos }^{8}x\\ & \\ & \gg \gg \sin x(1-2{\cos }^{2}x+{\cos }^{4}x)\cdot {\cos }^{8}x\\ & \\ & \gg \gg \sin x({\cos }^{8}x-2{\cos }^{10}x+{\cos }^{12}x)\\ & \\ & \gg \gg \sin x{\cos }^{8}x-2\sin x{\cos }^{10}x+\sin x{\cos }^{12}x\end{array}$$

$u$-sub for power-one

If $m=1$ or $n=1$, perform $u$-substitution to do the integral.

The other trig power becomes a $u$ power; the power-one becomes $du$.

For example, using $u=\cos x$ and thus $du=-\sin xdx$ we can do:

$$\int \sin x{\cos }^{8}xdx\gg \gg \int -{\cos }^{8}x(-\sin xdx)\gg \gg -\int u^{8}du$$

By combining these tricks you can do any power product with at least one odd power! Make sure to leave a power-one from the odd power when swapping an even bunch.

Notice

Even powers: $1={\sin }^{0}x={\cos }^{0}x$. So the method works for $\int {\sin }^{3}xdx$ and similar.

Power-to-frequency conversion

Using these ‘power-to-frequency’ identities (maybe repeatedly):

$${\sin }^{2}x=\frac{1}{2}(1-\cos 2x),{\cos }^{2}x=\frac{1}{2}(1+\cos 2x)$$

change an even power (either type) into an odd power of cosine.

For example, consider the power product:

$${\sin }^{4}x\cdot {\cos }^{6}x$$

You can substitute appropriate powers of ${\sin }^{2}x=\frac{1}{2}(1-\cos 2x)$ and ${\cos }^{2}x=\frac{1}{2}(1+\cos 2x)$:

$$\begin{array}{rl}{\sin }^{4}x\cdot {\cos }^{6}x& \gg \gg ({\sin }^{2}x{)}^2\cdot ({\cos }^{2}x{)}^3\\ & \\ & \gg \gg (\frac{1}{2}(1-\cos 2x){)}^2\cdot (\frac{1}{2}(1+\cos 2x){)}^3\end{array}$$

By doing some annoying algebra, this expression can be expanded as a sum of smaller powers of $\cos 2x$:

$$\begin{array}{c}(\frac{1}{2}(1-\cos 2x){)}^2\cdot (\frac{1}{2}(1+\cos 2x){)}^3\\ \\ \gg \gg \frac{1}{32}(1+\cos (2x)-2{\cos }^2(2x)-2{\cos }^3(2x)+{\cos }^4(2x)+{\cos }^5(2x))\end{array}$$

Each of these terms can be integrated by repeating the same techniques.

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Illustration

Example - Power product - odd power

Power product - odd power

Compute the integral:

$$\int {\cos }^{2}x\cdot {\sin }^{5}xdx$$

Solution

(1) Swap over the even bunch:

Max even bunch leaving power-one is ${\sin }^{4}x$.

$${\sin }^{5}x\gg \gg \sin x({\sin }^{2}x{)}^2\gg \gg \sin x(1-{\cos }^{2}x{)}^2$$

Apply to ${\sin }^{5}x$ in the integrand:

$$\int {\cos }^{2}x\cdot {\sin }^{5}xdx\gg \gg \int {\cos }^{2}x\cdot \sin x(1-{\cos }^{2}x{)}^{2}dx$$

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(2) Perform $u$-substitution on the power-one integrand:

Set $u=\cos x$. Hence $du=\sin xdx$. Recognize this in the integrand and convert:

$$\begin{array}{rl}\int {\cos }^{2}x\cdot \sin x(1-{\cos }^{2}x{)}^{2}dx& \gg \gg \int {\cos }^{2}x\cdot \left(1{\cos }^{2}x{)}^2\right(\sin xdx)\\ & \\ & \gg \gg \int u^2\cdot (1-u^2{)}^{2}du\end{array}$$

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(3) Integrate using power rule:

$$\int u^2\cdot (1-u^2{)}^{2}du=\frac{1}{3}{u}^3-\frac{2}{5}{u}^5+\frac{1}{7}{u}^7+C$$

Insert definition $u=\cos x$:

$$\begin{array}{c}\gg \gg \int {\cos }^{2}x\cdot {\sin }^{5}xdx\gg \gg \int u^2\cdot (1-u^2{)}^{2}du\\ \\ \gg \gg \frac{1}{3}{\cos }^{3}x-\frac{2}{5}{\cos }^{5}x+\frac{1}{7}{\cos }^{7}x+C\end{array}$$ Link to original

Theory

Theory 2

Trig power product: $\tan /\sec$ or $\cot /\csc$

A $\tan /\sec$ power product has this form:

$$\int {\tan }^{m}x\cdot {\sec }^{n}xdx$$

A $\cot /\csc$ power product has this form:

$$\int {\cot }^{m}x\cdot {\csc }^{n}xdx$$

To integrate these, swap an even bunch using:

• ${\tan }^{2}x+1={\sec }^{2}x$

OR:

• ${\cot }^{2}x+1={\csc }^{2}x$

Or do $u$-substitution using:

• $u=\tan x⇝du={\sec }^{2}xdx$
• $u=\sec x⇝du=\sec x\tan xdx$

OR:

• $u=\cot x⇝du=-{\csc }^{2}xdx$
• $u=\csc x⇝du=-\csc u\cot udx$

Note

There is no simple “power-to-frequency conversion” for tan / sec !

We can modify the power-one technique to solve some of these. We need to swap over an even bunch from the odd power so that exactly the $du$ factor is left behind.

Considering all the possibilities, one sees that this method works when:

• ${\tan }^{m}x$ is an odd power (with some secants present!)
• ${\sec }^{n}x$ is an even power

Quite a few cases escape this method:

• Any $\int {\tan }^{m}xdx$ with no power of $\sec x$
• Any $\int {\tan }^{m}x\cdot {\sec }^{n}xdx$ for $m$ even and $n$ odd

These tricks don’t work for $\int \tan xdx$ or $\int \sec xdx$ or $\int {\tan }^{4}x{\sec }^{5}xdx$, among others.

Special integrals: tan and sec

We have:

$$\begin{array}{rl}\int \tan xdx& =\ln |\sec x|+C\\ & \\ \int \sec xdx& =\ln |\sec x+\tan x|+C\end{array}$$

Note

These integrals should be memorized individually.

Extra - Deriving special integrals: tan and sec

The first formula can be found by $u$-substitution, considering that $\tan x=\frac{\sin x}{\cos x}$.

The second formula can be derived by multiplying $\sec x$ by a special “$1$”, computing instead $\int \frac{\sec x(\sec x+\tan x)}{\sec x+\tan x}dx$ by expanding the numerator and doing $u$-sub on the denominator.

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Illustration

Example - Power product - tan and sec

Power product - tan and sec

Compute the integral:

$$\int {\tan }^{5}x\cdot {\sec }^{3}xdx$$

Solution

(1) Try $du={\sec }^{2}xdx$:

Factor $du$ out of the integrand:

$$\int {\tan }^{5}x\cdot {\sec }^{3}xdx\gg \gg \int {\tan }^{5}x\cdot \sec x\left({\sec }^{2}xdx\right)$$

We then must swap over remaining $\sec x$ into the $\tan x$ type.

Cannot do this because $\sec x$ has odd power. Need even to swap.

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(2) Try again: $du=\sec x\tan xdx$:

Factor $du$ out of the integrand:

$$\int {\tan }^{5}x\cdot {\sec }^{3}xdx\gg \gg \int {\tan }^{4}x\cdot {\sec }^{2}x\left(\sec x\tan xdx\right)$$

Swap remaining $\tan x$ into $\sec x$ type:

$$\begin{array}{c}\gg \gg \int ({\tan }^{2}x{)}^2\cdot {\sec }^{2}x(\sec x\tan xdx)\\ \\ \gg \gg \int ({\sec }^{2}x-1{)}^2\cdot {\sec }^{2}x(\sec x\tan xdx)\end{array}$$

Substitute $u=\sec x$ and $du=\sec x\tan xdx$:

$$\gg \gg \int (u^2-1{)}^2\cdot u^{2}du$$

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(3) Integrate in $u$ and convert back to $x$:

$$\begin{array}{c}\gg \gg \int u^6-2u^4+u^{2}du\\ \\ \gg \gg \frac{u^7}{7}-2\frac{u^5}{5}+\frac{u^3}{3}+C\\ \\ \gg \gg \frac{{\sec }^{7}x}{7}-2\frac{{\sec }^{5}x}{5}+\frac{{\sec }^{3}x}{3}+C\end{array}$$ Link to original

Trig substitution

Videos

Review Videos

Videos, Math Dr. Bob:

• Trig sub 1: Basics and $\int \frac{1}{\sqrt{36-x^2}}dx$ and $\int \frac{x}{36+x^2}dx$ and $\int \frac{1}{\sqrt{x^2-36}}dx$
• Trig sub 2: $\int \frac{dx}{(1+x^2{)}^{5/2}}$
• Trig sub 3: $\int \frac{x^2}{\sqrt{1-4x^2}}dx$
• Trig sub 4: $\int \sqrt{e^{2x}-1}dx$
• Trig sub 5: $\int \frac{\sqrt{4-36x^2}}{x^2}dx$

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Theory

Theory 1

Certain algebraic expressions have a secret meaning that comes from the Pythagorean Theorem. This meaning has a very simple expression in terms of trig functions of a certain angle.

For example, consider the integral:

$$\int \frac{1}{x^2\sqrt{x^2-9}}dx$$

Now consider this triangle:

The triangle determines the relation $x=3\sec \theta$, and it implies $\sqrt{x^2-9}=3\tan \theta$.

Now plug these into the integrand above:

$$\frac{1}{x^2\sqrt{x^2-9^2}}\gg \gg \frac{1}{9{\sec }^2\theta \cdot 3\tan \theta }$$

Considering that $dx=3\sec \theta \tan \theta d\theta$, we obtain a very reasonable trig integral:

$$\begin{array}{rl}\int \frac{1}{x^2\sqrt{x^2-9^2}}dx\gg \gg & \int \frac{3\sec \theta \tan \theta }{27{\sec }^2\theta \tan \theta }d\theta \\ & \\ \gg \gg & \frac{1}{9}\int \cos \theta d\theta \gg \gg \frac{1}{9}\sin \theta +C\end{array}$$

We must rewrite this in terms of $x$ using $x=3\sec \theta$ to finish the problem. We need to find $\sin \theta$ assuming that $\sec \theta =\frac{x}{3}$. To do this, refer back to the triangle to see that $\sin \theta =\frac{\sqrt{x^2-9}}{x}$. Plug this in for our final value of the integral:

$$\frac{1}{9}\sin \theta +C\gg \gg \frac{\sqrt{x^2-9}}{9x}+C$$

Here is the moral of the story:

Pythagorean expressions

Re-express the Pythagorean expression using a triangle and a trig substitution.

In this way, we are able to eliminate square roots of quadratics.

There are always three steps for these trig sub problems:

• (1) Identify the trig sub: find the sides of a triangle and relevant angle $\theta$.
• (2) Solve a trig integral (often a power product).
• (3) Refer back to the triangle to convert the answer back to $x$.

To speed up your solution process for these problems, memorize these three transformations:

(1)

$$\sqrt{a^2-x^2}\stackrel{x=a\sin \theta }{\gg \gg }\sqrt{a^2-a^2{\sin }^2\theta }=a\cos \theta \text{from}1-{\sin }^2\theta ={\cos }^2\theta$$

(2)

$$\sqrt{a^2+x^2}\stackrel{x=a\tan \theta }{\gg \gg }\sqrt{a^2+a^2{\tan }^2\theta }=a\sec \theta \text{from}1+{\tan }^2\theta ={\sec }^2\theta$$

(3)

$$\sqrt{x^2-a^2}\stackrel{x=a\sec \theta }{\gg \gg }\sqrt{a^2{\sec }^2\theta -a^2}=a\tan \theta \text{from}{\sec }^2\theta -1={\tan }^2\theta$$

For a more complex quadratic with linear and constant terms, you will need to first complete the square for the quadratic and then do the trig substitution.

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Illustration

Example - Trig sub in quadratic: completing the square

Trig sub in quadratic - completing the square

Compute the integral:

$$\int \frac{dx}{\sqrt{x^2-6x+11}}$$

Solution

(1) Complete the square to obtain Pythagorean form:

$$x^2-6x+{\left(\frac{-6}{2}\right)}^2\gg \gg x^2-6x+9=(x-3{)}^2$$

Add and subtract to get desired constant term:

$$\begin{array}{c}{x}^2-6x+11\gg \gg x^2-6x+9-9+11\\ \\ \gg \gg (x-3{)}^2+2\end{array}$$

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(2) Perform shift substitution:

Set $u=x-3$ as inside the square:

$$(x-3{)}^2+2=u^2+2$$

Infer $du=dx$. Plug into integrand:

$$\int \frac{dx}{\sqrt{x^2-6x+11}}\gg \gg \int \frac{du}{\sqrt{u^2+2}}$$

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(3) Trig sub with $\tan \theta$:

Use substitution $u=\sqrt{2}\tan \theta$. (From triangle or memorized tip.)

Infer $du=\sqrt{2}{\sec }^2\theta d\theta$. Plug in data:

$$\int \frac{du}{\sqrt{u^2+2}}\gg \gg \int \frac{{\sec }^2\theta }{\sec \theta }d\theta =\int \sec \theta d\theta$$

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(4) Integrate using ad hoc memorized formula:

$$\int \sec \theta d\theta \gg \gg \ln |\tan \theta +\sec \theta |+C$$

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(5) Convert trig back to $x$:

First in terms of $u$, referring to the triangle:

$$\tan \theta =\frac{u}{\sqrt{2}},\sec \theta =\frac{\sqrt{u^2+2}}{\sqrt{2}}$$

Then in terms of $x$ using $u=x-3$. Plug everything in:

$$\ln |\tan \theta +\sec \theta |+C\gg \gg \ln \left|\frac{x-3}{\sqrt{2}}+\frac{\sqrt{(x-3{)}^2+2}}{\sqrt{2}}\right|+C$$

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(6) Simplify using log rules:

$$\ln \frac{f(x)}{a}\gg \gg \ln f(x)-\ln a$$

The common denominator $\frac{1}{\sqrt{2}}$ can be pulled outside as $-\ln \sqrt{2}$.

The new term $-\ln \sqrt{2}$ can be “absorbed into the constant” (redefine $C$).

So we write our final answer thus:

$$\ln \left|x-3+\sqrt{(x-3{)}^2+2}\right|+C$$Link to original

Partial fractions

Videos

Review Videos

Videos, Math Dr. Bob:

• Partial Fractions 01 - Distinct Linear Factors
• Partial Fractions 02 - Repeated Linear Factors
• Partial Fractions 03 - Distinct Mixed Factors
• Partial Fractions 04 - Repeated Quadratic Factors
• Partial Fractions 05 - Composition with $e^x$

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Theory

Theory 1

A rational function is a ratio of polynomials, for example:

$$\frac{P(x)}{Q(x)}=\frac{5x^2+x-28}{x^3-4x^2+x+6}$$

Partial fraction decomposition

The partial fraction decomposition of a rational function is a way of writing it as a sum of simple terms, like this:

$$\frac{3x^3-5x^2-6x+20}{x^4-3x^3+4x}=-\frac{2}{x+1}+\frac{2}{(x-2{)}^2}+\frac{5}{x}$$

Allowed denominators:

• Linear, e.g. $x-a$, or linear power, e.g. $(x-a{)}^n$
• Quadratic, e.g. $x^2+bx+c$, or quadratic power, e.g. $(x^2+bx+c{)}^n$
  • Condition: quadratics must be irreducible. (No roots, i.e. $b^2<4c$.)

Allowed numerators: constant (over linear power) or linear (over quadratic power)

These are allowed as simple terms in partial fraction decompositions:

$$\frac{1}{x^2+1},\frac{2x+1}{x^2+5},\frac{7}{5x-8},\frac{1}{x},\frac{1}{x^3}$$

These are not allowed:

$$\frac{x}{x-1},\frac{x^3+2}{x^2+1},\frac{1}{x^2-1},\frac{1}{x(x-1)}$$

These are allowed, showing irreducible quadratic and higher powers:

$$\frac{x}{x^2+1},\frac{x^3+2x+1}{(x^2+2{)}^2}$$

In this example the numerator is linear and the denominator is quadratic and irreducible.

To create a partial fraction decomposition, follow these steps:

1. Check denominator degree is higher
   • Else do long division
2. Factor denominator completely (even using irrational roots)
3. Write the generic sum of partial fraction terms with their constants

Repeated factors – special treatment – incrementing powers

4. Solve for constants

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Illustration

Partial fractions with repeated factor

Partial fractions with repeated factor

Find the partial fraction decomposition:

$$\frac{3x-9}{x^3+3x^2-4}$$

Solution

(1) Check that denominator degree is lower. $✓$

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(2) Factor denominator:

Rational Roots Theorem: check for roots at $\pm 1$ and $\pm 2$ and $\pm 4$.

Discover that $x=+1$ is a root. Therefore divide by $x-1$:

$$\frac{x^3+3x^2-4}{x-1}\gg \gg x^2+4x+4$$

Factor again:

$$x^2+4x+4\gg \gg (x+2{)}^2$$

Final factored form:

$$x^3+3x^2-4=(x-1)(x+2{)}^2$$

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(3) Write the generic PFD:

$$\frac{3x-9}{(x-1)(x+2{)}^2}=\frac{A}{x-1}+\frac{B}{x+2}+\frac{C}{(x+2{)}^2}$$

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(4) Solve for $A$, $B$, and $C$:

Multiply across by the common denominator:

$$3x-9=A(x+2{)}^2+B(x-1)(x+2)+C(x-1)$$

For $A$, set $x=1$, obtain:

$$\begin{array}{rlrl}& & 3\cdot 1-9& =A(1+2{)}^2+B\cdot 0+C\cdot 0\\ \gg \gg & & -6& =9A\\ \gg \gg & & A& =-2/3\end{array}$$

For $C$, set $x=-2$, obtain:

$$\begin{array}{rlrl}& & 3\cdot (-2)-9& =A\cdot 0+B\cdot 0+C\cdot (-3)\\ \gg \gg & & -15& =-3C\\ \gg \gg & & C& =5\end{array}$$

For $B$, insert prior results and solve.

Plug in $A$ and $C$:

$$3x-9=-\frac{2}{3}(x+2{)}^2+B(x-1)(x+2)+5(x-1)$$

Now plug in another convenient $x$, say $x=3$:

$$\begin{array}{rl}0& =-\frac{2}{3}\cdot 5^2+B\cdot 2\cdot 5+5\cdot 2\\ \frac{50}{3}-10& =10B\gg \gg B=\frac{2}{3}\end{array}$$

---

(4) Plug in $A$, $B$, $C$ for the final answer:

$$\frac{3x-9}{x^3+3x^2-4}=\frac{-2/3}{x-1}+\frac{2/3}{x+2}+\frac{5}{(x+2{)}^2}$$ Link to original

Theory

Theory 2

Partial fractions can be integrated using just a few techniques. Consider these terms:

$$\begin{array}{c}\frac{A}{x-a},\frac{A}{(x-a{)}^2},\frac{A}{(x-a{)}^3},\dots ,\\ \\ \text{and}\frac{A}{x^2+h^2},\text{and}\frac{Ax+B}{x^2+h^2}\end{array}$$

Linear power bottom

In order to integrate terms like this:

$$\frac{A}{(x-a{)}^n}$$

If $n=1$ then use log:

$$\int \frac{A}{x-a}dx=A\ln |x-a|+C$$

If $n>1$ then use power rule:

$$\int \frac{A}{(x-a{)}^n}dx\gg \gg \int A(x-a{)}^{-n}dx\gg \gg A\frac{(x-a{)}^{-n+1}}{-n+1}+C$$

Quadratic bottom, constant top

Formula for simple irreducible quadratics:

$$\int \frac{dx}{x^2+h^2}=\frac{1}{h}{\tan }^{-1}\left(\frac{x}{h}\right)+C$$

Memorize this formula!

Quadratic bottom, linear top

In order to integrate terms like this:

$$\frac{Ax+B}{x^2+h^2}$$

Break into separate terms:

$$\frac{Ax+B}{x^2+h^2}\gg \gg \frac{Ax}{x^2+h^2}+\frac{B}{x^2+h^2}$$

Then:

• First term with $x$ in top:

  $$\int \frac{Ax}{x^2+h^2}dx\gg \gg \frac{A}{2}\ln \left|x^2+h^2\right|+C$$

• Second term lacking $x$ in top:

  $$\int \frac{B}{x^2+h^2}dx\gg \gg \frac{B}{h}{\tan }^{-1}\left(\frac{x}{h}\right)+C$$

Extra - Completing the square when “no real roots”

To integrate terms with more general quadratics, like this:

$$\frac{A}{x^2+bx+c}$$

we need $b^2-4c<0$, i.e. “no real roots” of the quadratic. If that holds, then we can complete the square and substitute $u=\tan \theta$ as follows.

Look what happens when completing the square:

$$x^2+bx+c\gg \gg {\left(x+\frac{b}{2}\right)}^2-\frac{b^2}{4}+c$$

Notice that $b^2-4c<0$ is equivalent to the condition $-\frac{b^2}{4}+c>0$. Create a new label $Z=-\frac{b^2}{4}+c$. So this condition means $Z>0$ and we can safely define $\sqrt{Z}$.

Then a $u$-substitution $u=x+\frac{b}{2}$ simplifies the equation like this:

$$x^2+bx+c\gg \gg u^2+{\sqrt{Z}}^2$$

The quadratic formula $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$ shows that the condition $b^2-4c<0$ is equivalent to the condition “no real roots.” (In our case $a=1$. If we had $a\ne 1$, we could divide out this $a$ and change the others.)

So we see that “no real roots” is equivalent to the condition that the denominator can be converted to the format $x^2+h^2$ with some constant $h$.

At this point, to compute the integral, do trig sub with $u=\sqrt{Z}\tan \theta$ and $du=\sqrt{Z}{\sec }^2\theta d\theta$:

$$\begin{array}{c}\int \frac{Adx}{x^2+bx+c}\gg \gg \int \frac{A\sqrt{Z}{\sec }^2\theta d\theta }{Z{\sec }^2\theta }\\ \\ \gg \gg \frac{A}{\sqrt{Z}}\int d\theta \gg \gg \frac{A}{\sqrt{Z}}\theta +C\\ \\ \gg \gg \frac{A}{\sqrt{Z}}{\tan }^{-1}\left(\frac{x+b/2}{\sqrt{Z}}\right)+C\end{array}$$

Link to original

Illustration

Example - Repeated quadratic, linear tops

Partial fractions - repeated quadratic, linear tops

Compute the integral:

$$\int \frac{x^3+1}{(x^2+4{)}^2}dx$$

Solution

(1) Partial fraction decomposition:

• Numerator degree is lower than denominator.
• Factor denominator completely. (No real roots.)

Write generic PFD:

$$\frac{x^3+1}{(x^2+4{)}^2}=\frac{Ax+B}{x^2+4}+\frac{Cx+D}{(x^2+4{)}^2}$$

• Notice: repeated factor: use incrementing powers up to 2.
• Notice: linear over quadratic.

Common denominators and solve:

$$\begin{array}{c}{x}^3+1=(Ax+B)(x^2+4)+Cx+D\\ \\ \gg \gg x^3+1=A{x}^3+B{x}^2+(4A+C)x+4B+D\\ \\ \gg \gg A=1,B=0\\ \\ \gg \gg C=-4,D=1\end{array}$$

Therefore:

$$\frac{x^3+1}{(x^2+4{)}^2}=\frac{x}{x^2+4}+\frac{-4x+1}{(x^2+4{)}^2}$$

---

(2) Integrate:

Integrate the first term using substitution $u=x^2+4$:

$$\begin{array}{c}\int \frac{x}{x^2+4}dx\stackrel{u=x^2+4}{\gg \gg }\frac{1}{2}\int \frac{du}{u}\\ \\ \gg \gg \frac{1}{2}\ln |u|+C\gg \gg \frac{1}{2}\ln \left|x^2+4\right|+C\end{array}$$

Break up the second term:

$$\frac{-4x+1}{(x^2+4{)}^2}\gg \gg \frac{-4x}{(x^2+4{)}^2}+\frac{1}{(x^2+4{)}^2}$$

Integrate the first term of RHS:

$$\begin{array}{c}\int \frac{-4x}{(x^2+4{)}^2}dx\gg \gg -2\int \frac{du}{u^2}\\ \\ \gg \gg \frac{2}{u}+C\gg \gg \frac{2}{x^2+4}+C\end{array}$$

Integrate the second term of RHS using $x=2\tan \theta$:

$$\begin{array}{c}\int \frac{dx}{(x^2+4{)}^2}\gg \gg \int \frac{2{\sec }^2\theta d\theta }{16{\sec }^4\theta }\\ \\ \gg \gg \frac{1}{8}\int {\cos }^2\theta d\theta \\ \\ \gg \gg \frac{1}{8}\int \frac{1}{2}(1+\cos (2\theta ))d\theta \\ \\ \gg \gg \frac{1}{16}\theta +\frac{1}{32}\sin (2\theta )+C\\ \\ \gg \gg \frac{1}{16}{\tan }^{-1}\left(\frac{x}{2}\right)+\frac{1}{32}2\sin \theta \cos \theta +C\\ \\ \gg \gg \frac{1}{16}{\tan }^{-1}\left(\frac{x}{2}\right)+\frac{1}{32}2\frac{x}{\sqrt{x^2+2^2}}\frac{2}{\sqrt{x^2+2^2}}+C\\ \\ \gg \gg \frac{1}{16}{\tan }^{-1}\left(\frac{x}{2}\right)+\frac{1}{8}\frac{x}{x^2+2^2}+C\end{array}$$Link to original

Extra - “Rationalize a quotient” - convert into PFD

Extra - “Rationalize a quotient” - convert into PFD

Sometimes an integrand may be converted to a rational function using a substitution.

Consider this integral:

$$\int \frac{\sqrt{x+4}}{x}dx$$

Set $u=\sqrt{x+4}$, so $x=u^2-4$ and $dx=2udu$:

$$\gg \gg \int \frac{2udu}{u^2-4}$$

Now this rational function has a partial fraction decomposition:

$$\frac{2u}{u^2-4}\gg \gg \frac{2u}{(u-2)(u+2)}\gg \gg \frac{1}{u-2}+\frac{1}{u+2}$$

It is easy to integrate from there!

Practice exercises:

• To compute $\int \frac{\sqrt{x}}{x-1}dx$, try the substitution $u=\sqrt{x}$.
• To compute $\int \frac{dx}{\sqrt[2]{x}-\sqrt[3]{x}}$, try the substitution $u=\sqrt[6]{x}$.
• To compute $\int \frac{1}{x-\sqrt{x+2}}dx$, try the substitution $u=\sqrt{x+2}$.

Link to original

Simpson’s Rule

Videos

Review Videos

Videos, Math Dr. Bob:

• Simpson’s Rule with Error Bound

Videos, Organic Chemistry Tutor:

• Midpoint Rule and Riemans Sums
• Simpson’s Rule and Numerical Integration

Link to original

Theory - review

Theory 1

The Trapezoid Rule is a technique to approximate the area under a curve as the sum of areas of thin trapezoids whose top corners lie on the curve.

The tops of the trapezoids are lines that approximate the curve. They are determined as lines that agree with the curve at two points.

Trapezoid rule - area formula

Given a function $f$ and a partition of the range $[a,b]$ labeled by $x_0,x_1,\dots ,x_n$ (with $x_0=a$ and $x_n=b$), the Trapezoid Rule determines the area formula:

$$T_n=\frac{1}{2}\Delta x(y_0+2y_1+2y_2+\cdots +2y_{n-1}+y_n)$$

Notice the pattern in $2$s and see how this formula comes about:
The area of one trapezoid is $\Delta x\left(\frac{y_{j-1}+y_j}{2}\right)$. All vertical values $y_1,\dots ,y_{n-1}$ (excepting the endpoints $f(a)$ and $f(b)$) are represented in two trapezoids, so their contribution is doubled.

Extra - Trapezoid rule - error bound

The error of the Trapezoid Rule approximation is bounded by this formula:

$$\mathrm{Error}(T_n)\le \frac{K_2(b-a{)}^3}{12n^2}$$

Here $K_2$ is any number satisfying $K_2\ge |f^{\prime \prime }(x)|$ for $x\in [a,b]$.

The Midpoint Rule is a technique to approximate the area under a curve as the sum of areas of thin rectangles whose top midpoints lies on the curve.

The very same formula also represents the areas of trapezoids whose top midpoints lie on the curve and whose top line is tangent to the curve:

The reason they are equal is simple: when pivoting the top line on the ‘attached’ midpoint, the area of the trapezoid does not change.

Midpoint Rule - area formula

Given a function $f$ and a partition of the range $[a,b]$ labeled by $x_0,x_1,\dots ,x_n$ (with $x_0=a$ and $x_n=b$), the Midpoint Rule determines the area formula:

$$M_n=\Delta x(f(c_1)+f(c_2)+\cdots +f(c_{n-1})+f(c_n))$$

Here each $c_i$ is the midpoint of the interval $[x_{i-1},x_i]$. It can be given by the formula $c_i=a+(i-1/2)\Delta x$.

Extra - Midpoint Rule - error bound

The error of the Midpoint Rule approximation is bounded by this formula:

$$\mathrm{Error}(M_n)\le \frac{K_2(b-a{)}^3}{24n^2}$$

Here $K_2$ is any number satisfying $K_2\ge |f^{\prime \prime }(x)|$ for $x\in [a,b]$.

Notice that $M_n$ has an error bound that is $1/2$ of the bound for $T_n$. This does not mean that $M_n$ always has a smaller error than $T_n$. It means that without calculating the error, simply plugging numbers into the error bound formulas, we obtain a smaller bound for $M_n$ than for $T_n$. This is about our knowledge of the error, not the reality of the error.

Link to original

Theory

Theory 2

It turns out that the Midpoint Rule and the Trapezoid Rule tend to differ from the exact integral in opposite directions, and the Midpoint Rule tends to be twice as accurate. Therefore we may improve on both of them by constructing a weighted average of the formulas. This is called Simpson’s Rule.

Simpson’s Rule - defining formula

Simpson’s Rule is given by the weighted sum of the Trapezoid and Midpoint Rules:

$$S_n=\frac{1}{2}{T}_n+\frac{2}{3}{M}_n$$

Simpson’s Rule - computing formula

Given a function $f$ and a partition of the range $[a,b]$ labeled by $x_0,x_1,\dots ,x_n$ (with $x_0=a$ and $x_n=b$), Simpson’s Rule determines the area formula:

$$S_n=\frac{1}{3}\Delta x(y_0+4y_1+2y_2+4y_3+2y_4+\cdots +2y_{n-2}+4y_{n-1}+y_n)$$

Simpson’s Coefficient Pattern

Memorize the pattern for Simpson’s Rule:

$$1,4,2,4,2,4,2,\dots ,1$$

Simpson’s Rule - error bound

The error of Simpson’s Rule approximation is bounded by this formula:

$$\mathrm{Error}(S_n)\le \frac{K_4(b-a{)}^5}{180n^4}$$

Here $K_4$ is any number satisfying $K_4\ge |f^{(4)}(x)|$ for $x\in [a,b]$.

Simpson’s Rule $=$ “Parabola Rule”

The formula of Simpson’s Rule can also be explained or defined geometrically: it is the formula giving the sum of areas under small parabolas that meet the curve in three points.

There is a unique parabola passing through any three points with differing $x$-values:

These may be pieced together to form an approximation to the curve:

The area under the parabola through $P_0$, $P_1$, and $P_2$ is given by this formula:

$$\frac{h}{3}(y_0+4y_1+y_2)$$

This formula may be verified using basic calculus (area under a parabola) and a lot of algebra. (Ambitious students should derive it.)

The area under the parabola through $P_2$, $P_3$, and $P_4$ is given by a similar formula:

$$\frac{h}{3}(y_2+4y_3+y_4)$$

The Simpson’s Rule formula is the sum of all these formulas! So the $2$s in Simpson’s come from duplication of endpoint terms as the “rectangular” regions are stacked end-to-end.

Link to original

Illustration

Example - Simpson’s Rule on the Gaussian Distribution

Simpson’s Rule on the Gaussian distribution

The function $e^{x^2}$ is very important for probability and statistics, but it cannot be integrated analytically.

Apply Simpson’s Rule to approximate the integral:

$${\int }_0^1{e}^{x^2}dx$$

with $\Delta x=0.1$ and $n=10$. What error bound is guaranteed for this approximation?

Solution

(1) We need a table of values of $x_i$ and $y_i=f(x_i)$:

$x_i:$	$0.0$	$0.1$	$0.2$	$0.3$	$0.4$	$0.5$	$0.6$	$0.7$	$0.8$	$0.9$	$1.0$
$f(x_i):$	$e^{0.0^2}$	$e^{0.1^2}$	$e^{0.2^2}$	$e^{0.3^2}$	$e^{0.4^2}$	$e^{0.5^2}$	$e^{0.6^2}$	$e^{0.7^2}$	$e^{0.8^2}$	$e^{0.9^2}$	$e^{1.0^2}$
$\approx$	$1.000$	$1.010$	$1.041$	$1.094$	$1.174$	$1.284$	$1.433$	$1.632$	$1.896$	$2.248$	$2.718$

These can be plugged into the Simpson Rule formula to obtain our desired approximation:

$$\begin{array}{c}{S}_{10}=\frac{1}{3}\cdot 0.1\cdot (1.000+4\cdot 1.010+2\cdot 1.041+4\cdot 1.094+\cdots +2\cdot 1.896+4\cdot 2.248+2.718)\\ \\ \approx 1.463\end{array}$$

To find the error bound we need to find the smallest number we can manage for $K_4$.

Take four derivatives and simplify:

$$f^{(4)}(x)=(12+48x^2+16x^4)e^{x^2}$$

On the interval $x\in [0,1]$, this function is maximized at $x=1$. Use that for the optimal $K_4$:

$$f^{(4)}(1.000)=206.589$$

Finally we plug this into the error bound formula:

$$\frac{K_4(b-a{)}^5}{180n^4}=\frac{206.589\cdot 1.000^5}{180\cdot 10^4}\approx 0.0001$$ $$\gg \gg \mathrm{Error}(S_{10})\le 0.0001$$Link to original

Improper integrals

Videos

Review Videos

Videos, Math Dr. Bob:

• Improper integrals: Infinite limits
• Improper integrals: Vertical asymptote
• Improper integrals: $\int \frac{1}{x^p}dx$
• Improper integrals: $\int \frac{1}{x^2}{e}^{-1/x}dx$

Link to original

Theory

Theory 1

Improper integrals are those for which either a bound or the integrand itself become infinite somewhere on the interval of integration.

Examples:

$$(a){\int }_1^{\infty }\frac{1}{x^2}dx,(b){\int }_0^2\frac{1}{x}dx,(c){\int }_{-1}^{+1}\frac{1}{x^2}dx$$

(a) the upper bound is $\infty$
(b) the integrand goes to $\infty$ as $x\to 0^{+}$
(c) the integrand is $\infty$ at the point $0\in [-1,1]$

The limit interpretation of (a) is this:

$${\int }_1^{\infty }\frac{1}{x^2}dx=\underset{R\to \infty }{\lim }{\int }_1^R\frac{1}{x^2}dx$$

The limit interpretation of (b) is this:

$${\int }_0^2\frac{1}{x}dx=\underset{R\to 0^{+}}{\lim }{\int }_R^2\frac{1}{x}dx$$

The limit interpretation of (c) is this:

$$\begin{array}{c}{\int }_{-1}^{+1}\frac{1}{x^2}dx={\int }_{-1}^0\frac{1}{x^2}dx+{\int }_0^{+1}\frac{1}{x^2}dx\\ \\ =\underset{R\to 0^{-}}{\lim }{\int }_{-1}^R\frac{1}{x^2}dx+\underset{R\to 0^{+}}{\lim }{\int }_R^{+1}\frac{1}{x^2}dx\end{array}$$

These limits are evaluated using the usual methods.

An improper integral is said to be convergent or divergent according to whether it may be assigned a finite value through the appropriate limit interpretation.

For example, $(a)$ converges while $(b)$ diverges.

Link to original

Illustration

Example - Improper integral - infinite bound

Improper integral - infinite bound

Show that the improper integral ${\int }_2^{\infty }\frac{dx}{x^3}$ converges. What is its value?

Solution

(1) Improper integral definition:

$${\int }_2^{\infty }\frac{dx}{x^3}\gg \gg \underset{R\to \infty }{\lim }{\int }_2^R\frac{dx}{x^3}$$

---

(2) Replace infinity with a new symbol $R$:

$${\int }_2^R\frac{dx}{x^3}\gg \gg -\frac{1}{2}{x}^{-2}{|}_2^R\gg \gg \frac{1}{8}-\frac{1}{2R^2}$$

---

(3) Take limit as $R\to \infty$:

$$\begin{array}{c}\underset{R\to \infty }{\lim }{\int }_2^R\frac{dx}{x^3}\gg \gg \underset{R\to \infty }{\lim }\frac{1}{8}-\frac{1}{2R^2}\gg \gg \frac{1}{8}\\ \\ \gg \gg \text{converges}\end{array}$$ Link to original

Improper integral - infinite integrand

Improper integral - infinite integrand

Show that the improper integral ${\int }_0^9\frac{dx}{\sqrt{x}}$ converges. What is its value?

Solution

(1) Improper integral definition:

$${\int }_0^9\frac{dx}{\sqrt{x}}\gg \gg \underset{R\to 0^{+}}{\lim }{\int }_R^9\frac{dx}{\sqrt{x}}$$

---

(2) Switch $0$ to $R$ and integrate:

$$\begin{array}{c}{\int }_R^9\frac{dx}{\sqrt{x}}\gg \gg {\int }_R^9{x}^{-1/2}dx\\ \\ \gg \gg 2x^{+1/2}{|}_R^9\gg \gg 6-2\sqrt{R}\end{array}$$

---

(3) Take limit as $R\to 0^{+}$:

$$\begin{array}{c}\underset{R\to 0^{+}}{\lim }{\int }_R^9\frac{dx}{\sqrt{x}}\gg \gg \underset{R\to 0^{+}}{\lim }6-2\sqrt{R}\gg \gg 6\\ \\ \gg \gg \text{converges}\end{array}$$ Link to original

Example - Improper integral - infinity inside the interval

Example - Improper integral - infinity inside the interval

Does the integral ${\int }_{-1}^{+1}\frac{1}{x}dx$ converge or diverge?

Solution

WRONG APPROACH:

It is tempting to compute the integral incorrectly, like this:

$${\int }_{-1}^{+1}\frac{1}{x}dx\gg \gg \ln |x|{|}_{-1}^{+1}\gg \gg \ln |2|-\ln |-2|=0$$

But this is wrong! There is an infinite integrand at $x=0$. We must break it into parts!

---

RIGHT APPROACH:

(1) Break apart at $x=0$ discontinuity:

$${\int }_{-1}^{+1}\frac{1}{x}dx\gg \gg {\int }_{-1}^0\frac{1}{x}dx+{\int }_0^{+1}\frac{1}{x}dx$$

---

(2) Improper integral definition, both terms separately:

$$\gg \gg \underset{R\to 0^{-}}{\lim }{\int }_{-1}^R\frac{1}{x}dx+\underset{R\to 0^{+}}{\lim }{\int }_R^{+1}\frac{1}{x}dx$$

---

(3) Integrate:

$$\begin{array}{c}{\int }_{-1}^R\frac{1}{x}dx\gg \gg \ln |R|-\ln |-1|\gg \gg \ln |R|\\ \\ {\int }_R^{+1}\frac{1}{x}dx\gg \gg \ln |1|-\ln |R|\gg \gg -\ln R\end{array}$$

---

(4) Limits:

$$\begin{array}{c}\underset{R\to 0^{-}}{\lim }\ln |R|=-\infty ,\underset{R\to 0^{+}}{\lim }-\ln R=+\infty \\ \\ \gg \gg \text{diverges}\end{array}$$

Neither limit is finite. For ${\int }_{-1}^{+1}\frac{1}{x}dx$ to exist we’d need both of these limits to be finite. So the original integral diverges.

Link to original

Theory

Theory 2

Two tools allow us to determine convergence of a large variety of integrals. They are the comparison test and the $p$-integral cases.

Comparison test - integrals

The comparison test says:

• When an improper integral converges, every smaller integral converges.
• When an improper integral diverges, every bigger integral diverges.

Here, smaller and bigger are comparisons of the integrand at all values (accounting properly for signs), and the bounds are assumed to be the same.

For example, ${\int }_2^{\infty }\frac{dx}{x^3}$ converges, and $x^4>x^3$ implies $\frac{1}{x^4}<\frac{1}{x^3}$ (when $x>1$), therefore the comparison test implies that ${\int }_2^{\infty }\frac{dx}{x^4}$ converges.

$p$-integral cases

Assume $p>0$ and $a>0$. We have:

$$\begin{array}{rlrl}p>1:& {\int }_a^{\infty }\frac{dx}{x^p}\text{converges}& \text{and}& {\int }_0^a\frac{dx}{x^p}\text{diverges}\\ & & & \\ p<1:& {\int }_a^{\infty }\frac{dx}{x^p}\text{diverges}& \text{and}& {\int }_0^a\frac{dx}{x^p}\text{converges}\\ & & & \\ p=1:& {\int }_a^{\infty }\frac{dx}{x}\text{diverges}& \text{and}& {\int }_0^a\frac{dx}{x}\text{diverges}\end{array}$$

Proving the $p$-integral cases

It is easy to prove the convergence / divergence of each $p$-integral case using the limit interpretation and the power rule for integrals. (Or for $p=1$, using $\int \frac{1}{x}dx=\ln x+C$.)

Additional improper integral types

The improper integral ${\int }_{-\infty }^{a}f(x)dx$ also has a limit interpretation:

$${\int }_{-\infty }^{a}f(x)dx=\underset{R\to -\infty }{\lim }{\int }_R^{a}f(x)dx$$

The double improper integral ${\int }_{-\infty }^{\infty }f(x)dx$ has this limit interpretation:

$${\int }_{-\infty }^{\infty }f(x)dx=\underset{R\to -\infty }{\lim }{\int }_R^{a}f(x)dx+\underset{R\to \infty }{\lim }{\int }_a^{R}f(x)dx$$

Where $a$ is any finite number. This double integral does not exist if either limit does not exist for any value of $a$.

Double improper is not simultaneous!

Watch out! This may happen:

$${\int }_{-\infty }^{\infty }f(x)dx\ne \underset{R\to \infty }{\lim }{\int }_{-R}^{R}f(x)dx$$

This simultaneous limit might exist only because of internal cancellation in a case where the separate individual limits do not exist! We do not say ‘convergent’ in these cases!

Link to original

Illustration

Example - Comparison to $p$-integrals

Comparison to p-integrals

Determine whether the integral converges:

(a) ${\int }_2^{\infty }\frac{x^3}{x^4-1}dx$

(b) ${\int }_1^{\infty }\frac{1}{x^2+x+1}dx$

Solution

(a)
(1) Observe large $x$ tendency:

Consider large values. Notice the integrand tends toward $1/x$ for large $x$.

$$\frac{x^3}{x^4-1}⟶\frac{x^3}{x^4}\text{for}x\to \infty ,\text{and}\frac{x^3}{x^4}=\frac{1}{x}$$

---

(2) Try comparison to $1/x$:

$$\frac{x^3}{x^4-1}\stackrel{?}{>}\frac{1}{x}$$

Validate. Notice $x^4-1>0$ and $x>0$ when $x\ge 2$.

$$\begin{array}{c}\frac{x^3}{x^4-1}\stackrel{?}{>}\frac{1}{x}\\ \\ \gg \gg x^3\cdot x\stackrel{?}{>}1\cdot (x^4-1)\gg \gg x^4\check{>}{x}^4-1\end{array}$$

---

(3) Apply comparison test:

We know:

$$\frac{x^3}{x^4-1}>\frac{1}{x},{\int }_2^{\infty }\frac{1}{x}dx\text{diverges}$$

We conclude:

$${\int }_2^{\infty }\frac{x^3}{x^4-1}dx\text{diverges}$$

(b)
(1) Observe large $x$ tendency:

Consider large values. Notice the integrand tends toward $1/x^2$ for large $x$.

$$\frac{1}{x^2+x+1}⟶\frac{1}{x^2}\text{for}x\to \infty$$

---

(2) Try comparison to $1/x^2$:

$$\frac{1}{x^2+x+1}\stackrel{?}{<}\frac{1}{x^2}$$

Validate. Notice $x^2+x+1>0$ and $x^2>0$ when $x\ge 1$.

$$\begin{array}{c}\frac{1}{x^2+x+1}\stackrel{?}{<}\frac{1}{x^2}\\ \\ \gg \gg 1\cdot x^2\stackrel{?}{<}1\cdot (x^2+x+1)\gg \gg x^2\check{<}{x}^2+x+1\end{array}$$

---

(3) Apply comparison test:

We know:

$$\frac{1}{x^2+x+1}<\frac{1}{x^2},{\int }_1^{\infty }\frac{1}{x^2}dx\text{converges}$$

We conclude:

$${\int }_1^{\infty }\frac{1}{x^2+x+1}dx\text{converges}$$Link to original