MAT137/Unit 09
Unit 09 · Term 2

Integration Methods

A toolkit for finding antiderivatives: reverse the chain rule (substitution), reverse the product rule (by parts), and break rational functions apart (partial fractions).

By the end you can
Why it matters & where it comes from
Why it exists

Unlike derivatives, antiderivatives have no mechanical recipe — they take pattern-matching and cleverness. These techniques are the standard moves. And some functions, it turns out, have no elementary antiderivative at all.

In the real world

The bell curve $e^{-x^2}$ — the most important function in statistics — has no elementary antiderivative, so its integral must be computed numerically. Every statistical table and every p-value traces back to that fact.

Story & open questions

Liouville proved in the 1830s that integrals like $\int e^{-x^2}$ genuinely cannot be written in elementary terms — a rare and surprising "this is impossible" theorem. In 1968 the Risch algorithm turned the question into an algorithmic one for broad classes of elementary functions, allowing computer algebra systems to decide many cases.

Choosing a method

Integration technique is pattern recognition plus verification. Always check the derivative of your answer; that habit catches sign errors, missing constants, and wrong substitutions quickly.

Decision guide
Use substitution when you see a composition and the derivative of the inside is also present.
Use integration by parts when the integrand is a product and differentiating one factor makes it simpler.
Use partial fractions for rational functions after ensuring the numerator degree is smaller than the denominator degree.
Use trigonometric substitution when a square root matches $a^2-x^2$, $a^2+x^2$, or $x^2-a^2$.

Definite integral discipline

For definite integrals, every symbol must describe the same variable. After substituting $u=g(x)$, either change the bounds immediately or convert the antiderivative back to $x$ before evaluating.

Substitution

Rule

$\displaystyle\int f\!\big(g(x)\big)g'(x)\,dx=\int f(u)\,du$ with $u=g(x)$. For definite integrals, change the limits to $g(a)$ and $g(b)$.

Example

$\displaystyle\int 2x\,e^{x^2}\,dx$. Let $u=x^2$, $du=2x\,dx$: $\int e^u\,du=e^{x^2}+C$.

Integration by parts

Rule

$\displaystyle\int u\,dv=uv-\int v\,du.$ A good guide for choosing $u$ is LIATE (log, inverse-trig, algebraic, trig, exponential).

Example

$\displaystyle\int x e^{x}\,dx$. Take $u=x$, $dv=e^x dx$: $=xe^x-\int e^x\,dx=xe^x-e^x+C.$

Partial fractions

A proper rational function can be decomposed into simpler fractions. Repeated linear factors require a term for every power; irreducible quadratic factors require linear numerators, again with terms for repeated powers.

Example

$\displaystyle\int\frac{dx}{x^2-1}$. Write $\dfrac{1}{x^2-1}= \dfrac{1/2}{x-1}-\dfrac{1/2}{x+1}$, so the integral is $\tfrac12\ln\left|\dfrac{x-1}{x+1}\right|+C$.

Trigonometric substitution

For $\sqrt{a^2-x^2}$ use $x=a\sin\theta$; for $\sqrt{a^2+x^2}$ use $x=a\tan\theta$; for $\sqrt{x^2-a^2}$ use $x=a\sec\theta$.

Example

$\displaystyle\int\frac{dx}{\sqrt{1-x^2}}=\arcsin x+C$ — directly, or via $x=\sin\theta$, on an interval inside $(-1,1)$.

Common pitfalls

• After substitution, convert everything (including $dx$) to the new variable.
• In definite integrals, either change the limits or convert back before plugging in — don't mix.
• Partial fractions need a proper fraction first; if not, do polynomial division.

Practice

Problems adapted from the MAT137 question bank by Alfonso Gracia-Saz, used under CC BY-SA 4.0. Write full solutions before opening the answer.

Practice 1 · Substitution

Evaluate $\displaystyle\int x\cos(x^2)\,dx$.

Answer
$u=x^2$: $\tfrac12\sin(x^2)+C$.
Practice 2 · By parts

Evaluate $\displaystyle\int \ln x\,dx$.

Hint
By parts with $u=\ln x$, $dv=dx$: $x\ln x-x+C$.
Practice 3 · Partial fractions

Evaluate $\displaystyle\int\frac{3x+1}{x^2+x}\,dx$.

Hint
$\dfrac{3x+1}{x(x+1)}= \dfrac{1}{x}+\dfrac{2}{x+1}$, giving $\ln|x|+2\ln|x+1|+C$.
Practice 4 · By parts

Evaluate $\displaystyle\int_0^{1} x\,e^{-x}\,dx$.

Answer
By parts: $[-(x+1)e^{-x}]_0^1=1-2e^{-1}.$
Practice 5 · By parts twice

Evaluate $\displaystyle\int x^2\sin x\,dx$.

Answer
$-x^2\cos x+2x\sin x+2\cos x+C.$
Practice 6 · Power reduction

Evaluate $\displaystyle\int \cos^2 x\,dx$.

Hint
$\cos^2x=\tfrac{1+\cos2x}{2}$, so the integral is $\tfrac x2+\tfrac{\sin2x}{4}+C$.
Practice 7 · Trig substitution

Evaluate $\displaystyle\int_0^1\sqrt{1-x^2}\,dx$.

Answer
Let $x=\sin\theta$; the integral becomes $\int_0^{\pi/2}\cos^2\theta\,d\theta=\tfrac\pi4$ (a quarter disk).
Practice 8 · Partial fractions

Evaluate $\displaystyle\int\frac{1}{x^3-x}\,dx$.

Hint
$\dfrac1{x(x-1)(x+1)}= \dfrac{-1}{x}+\dfrac{1}{2(x-1)}+\dfrac{1}{2(x+1)}$, so the integral is $-\ln|x|+\tfrac12\ln|x-1|+\tfrac12\ln|x+1|+C.$

Expanded topic checklist

Course coverage. This unit is part of the complete MAT137 topic map. It should cover the following items before you move on:

Textbook/source alignment

Use the Alfonso Gracia-Saz MAT137 materials as the primary local spine, with the current UofT single-variable calculus topic split as the completeness check. The added items are original study prompts, not copied textbook exercises.

Core theorem map

The technique chain is: simplify algebra first, choose the derivative rule being reversed, then check the answer by differentiating.

Study routine. Make technique choice explicit: write one sentence saying why substitution, parts, partial fractions, or trig substitution is appropriate before computing.

Deep practice extensions

Use these after the main practice set. They are designed to force definition-level reasoning, theorem selection, and mixed application rather than routine calculation only.

Extension 1 · Substitution

Evaluate $\int x(1+x^2)^5\,dx$.

Guide
Let $u=1+x^2$, $du=2x dx$. The integral is $u^6/12+C=(1+x^2)^6/12+C$.
Extension 2 · Parts

Evaluate $\int x\ln x\,dx$ for $x>0$.

Guide
Take $u=\ln x$, $dv=x dx$. Then the integral is $(x^2/2)\ln x-\int x/2 dx=(x^2/2)\ln x-x^2/4+C$.
Extension 3 · Partial fractions

Decompose $1/(x^2-4)$ and integrate.

Guide
Use $1/(x^2-4)=A/(x-2)+B/(x+2)$ with $A=1/4$, $B=-1/4$. The integral is $(1/4)\ln|x-2|-(1/4)\ln|x+2|+C$.
Extension 4 · Check by differentiating

Why is differentiating your final answer a required final step in integration-technique problems?

Guide
The techniques involve substitutions, algebra, and constants where errors are easy. Differentiating confirms the antiderivative returns the original integrand on the intended domain.