When u-substitution doesn't work — typically when the integrand is a product of two different types of functions — integration by parts (IBP) is the go-to technique.
If $u$ and $v$ are differentiable functions, then:
$$\int u\,dv = uv - \int v\,du$$
In Leibniz notation: $\displaystyle\int u\,\frac{dv}{dx}\,dx = uv - \int v\,\frac{du}{dx}\,dx$
Derivation: From the product rule $\dfrac{d}{dx}[uv] = u\,v' + v\,u'$, integrate both sides and rearrange to get the IBP formula.
The hardest part of IBP is choosing which factor to call $u$ and which to call $dv$. The LIATE mnemonic gives the preferred order for $u$ (choose whichever type appears earliest in the list):
Example: For $\int x e^x\,dx$, $x$ is Algebraic (A) and $e^x$ is Exponential (E). Since A comes before E in LIATE, set $u = x$ and $dv = e^x\,dx$.
Evaluate $\displaystyle\int x e^x\,dx$.
Step 1 — Assign: $u = x$, $dv = e^x\,dx$
Step 2 — Differentiate / Integrate: $du = dx$, $v = e^x$
Step 3 — Apply formula:
$$\int x e^x\,dx = x e^x - \int e^x\,dx = x e^x - e^x + C = e^x(x-1) + C$$
Check: $\dfrac{d}{dx}[e^x(x-1)] = e^x(x-1) + e^x \cdot 1 = xe^x$ ✓
Evaluate $\displaystyle\int x\cos x\,dx$. (Hint: Let $u = x$, $dv = \cos x\,dx$)
Adjust the slider to see how $\int_0^a x e^x\,dx = e^a(a-1)+1$ grows. The shaded region equals $x e^x - e^x$ evaluated from 0 to $a$.
Figure 7.1 — Integration by Parts: $\displaystyle\int_0^a xe^x\,dx$
When the resulting $\int v\,du$ is still complex, apply IBP a second time. This often occurs with integrands like $x^2 e^x$ or $x^2 \sin x$.
Evaluate $\displaystyle\int x^2 e^x\,dx$.
Round 1: $u = x^2$, $dv = e^x\,dx$ ⟹ $du = 2x\,dx$, $v = e^x$
$$\int x^2 e^x\,dx = x^2 e^x - 2\int xe^x\,dx$$
Round 2: From Example 7.1, $\int xe^x\,dx = e^x(x-1) + C$
$$= x^2 e^x - 2e^x(x-1) + C = e^x(x^2 - 2x + 2) + C$$
For integrals like $\int e^x \sin x\,dx$, applying IBP twice regenerates the original integral. Let $I = \int e^x\sin x\,dx$. Applying IBP twice gives $I = -e^x\cos x + e^x\sin x - I$, so $2I = e^x(\sin x - \cos x)$, giving:
$$\int e^x \sin x\,dx = \frac{e^x(\sin x - \cos x)}{2} + C$$
AP Exam Tip: On the exam, IBP problems often come in the FRQ section requiring both the setup and evaluation. Always write your $u$, $dv$, $du$, $v$ assignments clearly — this earns partial credit even if you make an arithmetic error later. The LIATE rule works for most problems, but trust your judgment when the rule seems ambiguous.
Trigonometric integrals involve products or powers of trig functions. Strategies depend on whether the powers are odd or even.
(a) $\displaystyle\int \sin^3 x\,dx$ — odd power of sin
Write $\sin^3 x = \sin^2 x \cdot \sin x = (1-\cos^2 x)\sin x$. Let $u = \cos x$, $du = -\sin x\,dx$:
$$\int (1-u^2)(-du) = -u + \frac{u^3}{3} + C = -\cos x + \frac{\cos^3 x}{3} + C$$
(b) $\displaystyle\int \cos^2 x\,dx$ — even power, use half-angle
$$\int \frac{1+\cos 2x}{2}\,dx = \frac{x}{2} + \frac{\sin 2x}{4} + C$$
(c) $\displaystyle\int \sin^2 x\cos^2 x\,dx$ — both even
Use $\sin^2 x\cos^2 x = \dfrac{1}{4}\sin^2 2x = \dfrac{1}{4} \cdot \dfrac{1-\cos 4x}{2} = \dfrac{1-\cos 4x}{8}$:
$$= \frac{x}{8} - \frac{\sin 4x}{32} + C$$
Evaluate $\displaystyle\int \cos^3 x\,dx$.
Compare $\sin^2 x$ and $\cos^2 x$ using their half-angle forms. Note how both integrate to give $x/2$ over a full period — this reflects Pythagoras: $\sin^2 + \cos^2 = 1$.
Figure 7.2 — $\sin^2 x$ and $\cos^2 x$: half-angle identities
Partial fraction decomposition breaks a rational function into simpler pieces that are easier to integrate. It applies when the degree of the numerator is less than the degree of the denominator (if not, do polynomial long division first).
Forms:
Evaluate $\displaystyle\int \frac{3x+5}{(x-1)(x+2)}\,dx$.
Step 1 — Decompose: $\dfrac{3x+5}{(x-1)(x+2)} = \dfrac{A}{x-1} + \dfrac{B}{x+2}$
Step 2 — Multiply by $(x-1)(x+2)$: $3x+5 = A(x+2) + B(x-1)$
Step 3 — Solve for constants:
$x=1$: $8 = 3A \Rightarrow A = 8/3$
$x=-2$: $-1 = -3B \Rightarrow B = 1/3$
Step 4 — Integrate:
$$\int \frac{3x+5}{(x-1)(x+2)}\,dx = \frac{8}{3}\int\frac{dx}{x-1} + \frac{1}{3}\int\frac{dx}{x+2}$$
$$= \frac{8}{3}\ln|x-1| + \frac{1}{3}\ln|x+2| + C$$
Evaluate $\displaystyle\int \frac{x+4}{x(x+2)^2}\,dx$.
Decomposition: $\dfrac{x+4}{x(x+2)^2} = \dfrac{A}{x} + \dfrac{B}{x+2} + \dfrac{C}{(x+2)^2}$
Multiply through: $x+4 = A(x+2)^2 + Bx(x+2) + Cx$
$x=0$: $4 = 4A \Rightarrow A = 1$
$x=-2$: $2 = -2C \Rightarrow C = -1$
Expand and compare $x^2$ terms: $0 = A + B \Rightarrow B = -1$
$$\int\left(\frac{1}{x} - \frac{1}{x+2} - \frac{1}{(x+2)^2}\right)dx = \ln|x| - \ln|x+2| + \frac{1}{x+2} + C$$
Decompose and integrate: $\displaystyle\int \frac{2}{x^2-1}\,dx = \int \frac{2}{(x-1)(x+1)}\,dx$
AP Exam Tip: On AP Calc BC, partial fractions often appear in FRQ problems involving logistic growth or population models (where separation of variables leads to a rational integral). Always check: if the degree of the numerator $\geq$ degree of the denominator, perform polynomial long division first before setting up partial fractions.
Improper integrals are a BC-only topic. They may appear on the BC exam in multiple choice or FRQ format.
An improper integral has either an infinite limit of integration or an integrand that is unbounded (vertical asymptote) on the interval.
Type 1 (Infinite limits):
$$\int_a^\infty f(x)\,dx = \lim_{t\to\infty}\int_a^t f(x)\,dx$$
$$\int_{-\infty}^b f(x)\,dx = \lim_{t\to-\infty}\int_t^b f(x)\,dx$$
Type 2 (Vertical asymptote at $x=c$ inside $[a,b]$):
$$\int_a^b f(x)\,dx = \lim_{t\to c^-}\int_a^t f(x)\,dx + \lim_{t\to c^+}\int_t^b f(x)\,dx$$
The integral converges if the limit exists and is finite; otherwise it diverges.
(a) Converges: $\displaystyle\int_1^\infty \frac{1}{x^2}\,dx$
$$= \lim_{t\to\infty}\int_1^t x^{-2}\,dx = \lim_{t\to\infty}\left[-\frac{1}{x}\right]_1^t = \lim_{t\to\infty}\left(-\frac{1}{t} + 1\right) = 1$$
(b) Diverges: $\displaystyle\int_1^\infty \frac{1}{x}\,dx$
$$= \lim_{t\to\infty}\Big[\ln x\Big]_1^t = \lim_{t\to\infty}(\ln t - 0) = \infty \quad \text{(diverges)}$$
(c) Vertical asymptote: $\displaystyle\int_0^1 \frac{1}{\sqrt{x}}\,dx$
$$= \lim_{t\to 0^+}\int_t^1 x^{-1/2}\,dx = \lim_{t\to 0^+}\Big[2\sqrt{x}\Big]_t^1 = \lim_{t\to 0^+}(2 - 2\sqrt{t}) = 2$$
$$\int_1^\infty \frac{1}{x^p}\,dx \text{ converges if and only if } p > 1$$
Specifically: $\displaystyle\int_1^\infty \frac{1}{x^p}\,dx = \begin{cases} \dfrac{1}{p-1} & \text{if } p > 1 \\ \text{diverges} & \text{if } p \leq 1 \end{cases}$
Use the slider to change $p$ in $\int_1^\infty x^{-p}\,dx$. Watch how the shaded area (from 1 to $b=10$) grows unboundedly when $p \leq 1$ but stays bounded when $p > 1$.
Figure 7.3 — p-integral: convergence depends on $p > 1$
Determine whether $\displaystyle\int_0^\infty e^{-x}\,dx$ converges or diverges. If it converges, find its value.
With multiple techniques available, part of the skill is recognizing which to use:
| Integrand Type | Technique | Example |
|---|---|---|
| Composite function $f(g(x))g'(x)$ | u-substitution | $\int 2x\cos(x^2)\,dx$ |
| Product: poly × trig/exp | Integration by Parts (LIATE) | $\int x\sin x\,dx$ |
| Product: log/arctan × poly | IBP, set log/arctan as $u$ | $\int x\ln x\,dx$ |
| Powers of sin/cos | Trig identities + substitution | $\int \sin^4 x\,dx$ |
| Rational function | Partial fractions | $\int \frac{1}{x^2-4}\,dx$ |
| $\sqrt{a^2-x^2}$ type | Trig substitution ($x=a\sin\theta$) | $\int \sqrt{4-x^2}\,dx$ |
| Infinite limits or asymptote | Improper integral (limit) | $\int_0^\infty e^{-2x}\,dx$ |
Choose the best technique and set up (but don't complete) each integral:
(a) $\displaystyle\int \ln x\,dx$ → IBP: $u = \ln x$, $dv = dx$, so $du = \frac{1}{x}dx$, $v = x$; gives $x\ln x - \int x \cdot \frac{1}{x}\,dx = x\ln x - x + C$
(b) $\displaystyle\int \frac{x^2+1}{x^2-1}\,dx$ → Degree of numerator = degree of denominator, so long division first: $\dfrac{x^2+1}{x^2-1} = 1 + \dfrac{2}{x^2-1}$, then use partial fractions on $\dfrac{2}{(x-1)(x+1)}$.
(c) $\displaystyle\int \arctan x\,dx$ → IBP: $u = \arctan x$, $dv = dx$; gives $x\arctan x - \dfrac{1}{2}\ln(1+x^2) + C$
AP Exam Tip: On the AB exam, IBP and partial fractions are the most heavily tested techniques. For BC, improper integrals appear in both multiple choice and FRQ. A common MC trick: $\int_1^\infty \frac{1}{x}\,dx$ diverges (this surprises many students) while $\int_1^\infty \frac{1}{x^2}\,dx = 1$ converges. Memorize the p-integral cutoff: converges iff $p > 1$.
Evaluate $\displaystyle\int x e^{-x}\,dx$.
Evaluate $\displaystyle\int \ln x\,dx$.
Evaluate $\displaystyle\int \sin^2 x\,dx$.
Evaluate $\displaystyle\int \frac{5}{x^2+x-2}\,dx$.
Determine convergence and evaluate: $\displaystyle\int_2^\infty \frac{1}{x^3}\,dx$.
Evaluate $\displaystyle\int_0^1 \frac{1}{(x+1)(x+3)}\,dx$.
(BC) Determine if $\displaystyle\int_{-\infty}^{\infty} xe^{-x^2}\,dx$ converges. If so, find its value.
(AP Free Response Style) Let $R$ be the region bounded by $y = xe^{-x}$, the $x$-axis, and $x = 0$, extended to infinity.
(a) Find the area of $R$. (b) Write a definite integral and evaluate it.
$\displaystyle\int u\,dv = uv - \int v\,du$. Use LIATE to choose $u$: Logarithmic, Inverse trig, Algebraic, Trig, Exponential.
For $\sqrt{a^2-x^2}$: let $x = a\sin\theta$. For $\sqrt{a^2+x^2}$: let $x = a\tan\theta$. For $\sqrt{x^2-a^2}$: let $x = a\sec\theta$.
Decompose $\dfrac{P(x)}{Q(x)}$ into simpler fractions when $\deg P < \deg Q$ and $Q$ factors over the reals.
$\displaystyle\int_a^\infty f\,dx = \lim_{b\to\infty}\int_a^b f\,dx$. Converges if the limit exists; diverges otherwise.
$\displaystyle\int u\,dv = uv - \int v\,du$
$\sin^2\theta = \dfrac{1-\cos 2\theta}{2}$ $\cos^2\theta = \dfrac{1+\cos 2\theta}{2}$
$\sqrt{a^2-x^2}$: $x=a\sin\theta$ $\sqrt{a^2+x^2}$: $x=a\tan\theta$ $\sqrt{x^2-a^2}$: $x=a\sec\theta$
If $0 \leq f(x) \leq g(x)$ and $\int g$ converges, then $\int f$ converges. Contrapositive also holds.