Chapter 07

Medium

Techniques of Integration

00 · Symbol Glossary

$u,\ du$u-substitution variables

A new variable $u$ chosen to replace some expression inside the integrand, together with its differential $du$ . Distinct from Chapter 03's chain-rule $u$ , but built on the identical idea of naming an inner expression.

$\ln\lvert x\rvert$natural log of the absolute value of x

The antiderivative of $\dfrac{1}{x}$ — the exception flagged in Chapter 06 for $n=-1$ in the power rule. The absolute value keeps the expression defined for negative $x$ , since $\ln$ itself is only defined for positive inputs.

$\int u\,dv = uv - \int v\,du$integration by parts formula

A formula for integrating a product of two expressions, derived by reversing the product rule (Chapter 02). $u$ and $dv$ are chosen from the two pieces of the original integrand.

$\dfrac{A}{x-r_1} + \dfrac{B}{x-r_2}$partial fraction decomposition

Rewriting a single rational expression as a sum of simpler fractions, each with an antiderivative that is a natural logarithm — used to integrate rational functions the power rule cannot handle directly.

01 · u-Substitution

The power rule for integration (Chapter 06) only handles a bare power of $x$ . When the integrand is a composite function — the reverse of a chain-rule derivative from Chapter 03 — a substitution undoes that composition first.

Definition — u-Substitution

To evaluate $\int f(g(x))\,g'(x)\,dx$ :

Let $u = g(x)$ .
Compute $du = g'(x)\,dx$ .
Rewrite the integral entirely in terms of $u$ : $\int f(u)\,du$ .
Integrate with respect to $u$ , then substitute $g(x)$ back in for $u$ .

Step-by-step — Evaluate $\int 2x(x^2+1)^4\,dx$

Choose the inner expression as $u$ : let

u = x^2+1

— this is the expression being raised to a power, matching the inner-function role from Chapter 03's chain rule.

Differentiate to find $du$ :

\dfrac{du}{dx} = 2x \implies du = 2x\,dx

Substitute — notice $2x\,dx$ appears exactly in the integrand:

\int 2x(x^2+1)^4\,dx = \int u^4\,du

Integrate using the power rule (Chapter 06):

\int u^4\,du = \dfrac{u^5}{5}+C

Substitute $x^2+1$ back in for $u$ :

\dfrac{(x^2+1)^5}{5}+C

Verify by differentiating the result (Chapter 03's chain rule):

\dfrac{d}{dx}\left[\dfrac{(x^2+1)^5}{5}\right] = \dfrac{1}{5}\cdot5(x^2+1)^4\cdot2x = 2x(x^2+1)^4

— matches the original integrand. ✓

❌ Substituting when the exact du is not present

Attempting $\int x(x^2+1)^4\,dx$ (note: no factor of $2$ in front of $x$ , unlike the worked example) with $u=x^2+1$ , $du=2x\,dx$ .

Why it breaks: the integrand only has $x\,dx$ , not $2x\,dx$ — the substitution needs $du=2x\,dx$ to appear exactly, and $x\,dx$ is not literally $du$ .

Correct fix: solve for the missing piece: $x\,dx = \dfrac{1}{2}du$ (dividing the $du$ equation by $2$ ), then substitute: $\int x(x^2+1)^4dx = \int u^4\cdot\dfrac12\,du = \dfrac12\int u^4\,du = \dfrac{u^5}{10}+C = \dfrac{(x^2+1)^5}{10}+C$ .

Consequence: a missing constant factor is recoverable by algebra, but a substitution is only directly usable when every part of $du$ (up to a constant multiple) is present in the original integrand — if $x$ appeared with the wrong power or in a different combination, no constant fix would work, and a different technique would be needed.

02 · The Antiderivative of 1/x

Chapter 06 flagged $n=-1$ as the exception to the power rule for integration. Its antiderivative is the natural logarithm.

Definition — Antiderivative of 1/x

\int \frac{1}{x}\,dx = \ln\lvert x\rvert + C

Verify by differentiating: $\dfrac{d}{dx}\big[\ln\lvert x\rvert\big] = \dfrac{1}{x}$ for $x\neq0$ , matching Chapter 06's requirement $F'(x)=f(x)$ .

Step-by-step — Evaluate $\int \dfrac{3x^2}{x^3+2}\,dx$ using substitution

Choose $u$ : let

u = x^3+2

(the denominator).

Differentiate:

du = 3x^2\,dx

— matching the numerator exactly.

Substitute:

\int \dfrac{3x^2}{x^3+2}\,dx = \int \dfrac{1}{u}\,du

Integrate:

\int\dfrac1u\,du = \ln\lvert u\rvert + C

Substitute back:

\ln\lvert x^3+2\rvert + C

03 · Integration by Parts

u-substitution reverses the chain rule. Integration by parts reverses the product rule (Chapter 02), for integrands that are a product of two different types of expressions where substitution alone does not simplify things.

Definition — Integration by Parts

Starting from the product rule $\dfrac{d}{dx}[uv] = u'v+uv'$ , integrate both sides and rearrange:

\int u\,dv = uv - \int v\,du

$u$ — the factor chosen to be differentiated (should get simpler after differentiating).

$dv$ — the remaining factor, chosen to be integrated (should be straightforward to integrate).

Where the trig and exponential antiderivatives below come from

The worked examples in this section use $\sin x$ , $\cos x$ , and $e^x$ alongside polynomials. Their derivatives — $\dfrac{d}{dx}[\sin x]=\cos x$ , $\dfrac{d}{dx}[\cos x]=-\sin x$ , $\dfrac{d}{dx}[e^x]=e^x$ — were derived from the limit definition in Chapter 02, Sections 09–10. Reading each antiderivative below as "the function whose derivative is the integrand" reuses those established facts directly, run in reverse.

Choosing u wisely

A common heuristic: choose $u$ to be the piece that becomes simpler when differentiated — for instance, $x$ differentiates to $1$ , while an exponential or trigonometric factor tends to stay structurally similar under differentiation and is often easier to integrate. Picking the wrong piece for $u$ can produce an integral that is more complicated than the original.

Step-by-step — Evaluate $\int x\cos x\,dx$

Choose $u$ and $dv$ : let

u=x

(simplifies to

1

under differentiation) and

dv=\cos x\,dx

Compute $du$ and $v$ :

du = 1\,dx = dx

; integrating

dv

v=\sin x

(Chapter 02, Section 09:

\frac{d}{dx}[\sin x]=\cos x

, so

\sin x

is an antiderivative of

\cos x

Apply the formula:

\int x\cos x\,dx = uv - \int v\,du = x\sin x - \int \sin x\,dx

Evaluate the remaining integral:

\int\sin x\,dx = -\cos x + C

(since

\frac{d}{dx}[-\cos x]=\sin x

Assemble:

x\sin x - (-\cos x) + C = x\sin x + \cos x + C

Verify by differentiating (product rule + Chapter 02):

\dfrac{d}{dx}\big[x\sin x + \cos x\big] = (1\cdot\sin x + x\cos x) + (-\sin x) = x\cos x

. Matches. ✓

❌ Choosing u and dv the wrong way around

On the same integral $\int x\cos x\,dx$ , choosing instead $u=\cos x$ , $dv = x\,dx$ : then $du=-\sin x\,dx$ , $v=\dfrac{x^2}{2}$ .

Why it breaks: the formula gives $\int x\cos x\,dx = \dfrac{x^2}{2}\cos x - \int \dfrac{x^2}{2}(-\sin x)\,dx = \dfrac{x^2}{2}\cos x + \dfrac12\int x^2\sin x\,dx$ . The remaining integral $\int x^2\sin x\,dx$ is now more complicated than the original — the power of $x$ went up, not down.

Consequence: this choice does not fail outright (the formula is still valid), but it makes no progress toward a solvable integral. The heuristic in the note above — pick $u$ so differentiating it simplifies the expression — exists precisely to avoid this dead end.

04 · Partial Fractions

Rational functions (one polynomial divided by another) sometimes cannot be integrated directly, but if the denominator factors into distinct linear pieces, the fraction can be split into simpler pieces that each integrate to a logarithm (Section 02).

Definition — Partial Fraction Decomposition (distinct linear factors)

For a rational function with denominator $(x-r_1)(x-r_2)$ (distinct roots) and numerator of lower degree than the denominator:

\frac{p(x)}{(x-r_1)(x-r_2)} = \frac{A}{x-r_1} + \frac{B}{x-r_2}

for constants $A,B$ to be determined.

Step-by-step — Evaluate $\int \dfrac{5x-1}{(x-1)(x+2)}\,dx$

Set up the decomposition:

\dfrac{5x-1}{(x-1)(x+2)} = \dfrac{A}{x-1} + \dfrac{B}{x+2}

Clear denominators by multiplying both sides by $(x-1)(x+2)$ :

5x-1 = A(x+2) + B(x-1)

Solve for $A$ by choosing $x=1$ (this makes the $B$ term vanish):

5(1)-1 = A(1+2)+B(0) \implies 4 = 3A \implies A=\dfrac43

Solve for $B$ by choosing $x=-2$ (this makes the $A$ term vanish):

5(-2)-1 = A(0)+B(-2-1) \implies -11 = -3B \implies B=\dfrac{11}{3}

Rewrite the integral using the decomposition:

\int\left(\dfrac{4/3}{x-1} + \dfrac{11/3}{x+2}\right)dx

Integrate each term using Section 02's log rule:

\dfrac43\ln\lvert x-1\rvert + \dfrac{11}{3}\ln\lvert x+2\rvert + C

05 · Choosing a Technique

Definition — Technique Selection Checklist

Is the integrand a bare power of $x$ , or a sum of such powers? → power rule (Chapter 06).
Does an inner expression's derivative (up to a constant) appear elsewhere in the integrand? → u-substitution.
Is the integrand a product of two structurally different factors (polynomial times trig/exponential) where substitution does not simplify it? → integration by parts.
Is the integrand a rational function whose denominator factors into distinct linear pieces? → partial fractions.

06 · Exercises

EXERCISE 7.1

Look for an inner expression whose derivative (up to a constant multiple) appears elsewhere in the integrand.

$\int 4x^3(x^4+5)^6\,dx$ . Let $u=x^4+5$ , $du=4x^3\,dx$ — matches exactly. $\int u^6\,du = \dfrac{u^7}{7}+C = \dfrac{(x^4+5)^7}{7}+C$ .

Evaluate $\int 4x^3(x^4+5)^6\,dx$ using u-substitution.

EXERCISE 7.2

The numerator is not an exact match for $du$ — find the missing constant factor and adjust.

$\int \dfrac{x}{x^2+4}\,dx$ . Let $u=x^2+4$ , $du=2x\,dx \implies x\,dx=\frac12du$ . $\int\dfrac{1}{u}\cdot\frac12\,du = \frac12\ln\lvert u\rvert+C = \frac12\ln(x^2+4)+C$ (absolute value bars droppable since $x^2+4>0$ always).

Evaluate $\int \dfrac{x}{x^2+4}\,dx$ .

EXERCISE 7.3

Choose $u$ to be the factor that simplifies under differentiation, then apply the integration by parts formula once.

$\int x e^x\,dx$ . Let $u=x$ , $dv=e^x\,dx$ ; $du=dx$ , $v=e^x$ . $\int xe^x\,dx = xe^x - \int e^x\,dx = xe^x - e^x + C$ .

Evaluate $\int x e^x\,dx$ using integration by parts.

EXERCISE 7.4

Set up partial fractions with unknowns $A$ and $B$ , clear denominators, then substitute the roots of each factor to solve for each unknown separately.

$\int \dfrac{7x+3}{(x+1)(x-3)}\,dx$ . Decompose: $\dfrac{A}{x+1}+\dfrac{B}{x-3}$ . Clear: $7x+3=A(x-3)+B(x+1)$ . At $x=-1$ : $-4=A(-4) \implies A=1$ . At $x=3$ : $24=B(4) \implies B=6$ . Integral: $\ln\lvert x+1\rvert + 6\ln\lvert x-3\rvert + C$ .

Evaluate $\int \dfrac{7x+3}{(x+1)(x-3)}\,dx$ using partial fractions.

EXERCISE 7.5

Substitution needs the exact differential (up to a constant) present in the integrand. Check what power of $x$ would actually be needed for a clean substitution here.

$\int x^2(x^2+1)^4\,dx$ cannot be solved by substituting $u=x^2+1$ , since $du=2x\,dx$ requires exactly one power of $x$ outside the parentheses, but the integrand has $x^2$ — an extra, unmatched factor of $x$ remains after substitution that cannot be written purely in terms of $u$ (since $x^2 = u-1$ only accounts for one of the two factors, and the substitution $x^2\,dx$ still has a leftover $x\,dx$ vs $dx$ mismatch). Simple u-substitution fails here; the integral instead requires full binomial expansion of $(x^2+1)^4$ followed by term-by-term power-rule integration.

Explain why $\int x^2(x^2+1)^4\,dx$ cannot be evaluated with the substitution $u=x^2+1$ the way $\int 2x(x^2+1)^4\,dx$ could in Section 01. What must be done instead?

EXERCISE 7.6

Accumulated value over time is a definite integral of the rate function; if the rate involves a rational expression with a linear-factor denominator, partial fractions may be required before integrating.

Cash flow rate $r(t) = \dfrac{100}{(t+1)(t+3)}$ for $t\in[0,2]$ . Decompose: $\dfrac{A}{t+1}+\dfrac{B}{t+3}$ , with $100=A(t+3)+B(t+1)$ . At $t=-1$ : $100=2A\implies A=50$ . At $t=-3$ : $100=-2B\implies B=-50$ . Antiderivative: $F(t)=50\ln\lvert t+1\rvert - 50\ln\lvert t+3\rvert$ . $F(2)=50\ln3-50\ln5$ . $F(0)=50\ln1-50\ln3=0-50\ln3$ . Total $= F(2)-F(0) = 50\ln3-50\ln5-(-50\ln3)=100\ln3-50\ln5\approx100(1.0986)-50(1.6094)\approx109.86-80.47\approx29.39$ .

A cash flow rate is $r(t)=\dfrac{100}{(t+1)(t+3)}$ dollars per day, $t\in[0,2]$ . Find the total accumulated cash using partial fractions and FTC Part 2.

07 · Chapter Summary

Concept	Formula / Rule
u-Substitution	Reverses the chain rule; $u=g(x)$ , $du=g'(x)dx$
Antiderivative of $1/x$	$\int\frac1x\,dx = \ln\lvert x\rvert+C$
Integration by parts	$\int u\,dv = uv-\int v\,du$ ; reverses the product rule
Partial fractions	Split $\frac{p(x)}{(x-r_1)(x-r_2)}$ into $\frac{A}{x-r_1}+\frac{B}{x-r_2}$
Technique selection	Match integrand structure to substitution / by-parts / partial fractions

Next: Chapter 08 — Improper Integrals, which extends the definite integral to unbounded intervals and unbounded integrands.