Practice Problems – Techniques of Integration – AP Calculus

Below are 20 practice problems covering all topics from Module 03B: Techniques of Integration. Topics include: substitution, integration by parts, polynomial long division, partial fractions, trigonometric substitution, and improper integrals. Each problem includes a hidden solution — click the button to reveal the step-by-step answer.

Problem 01 Substitution

Find $\displaystyle \int 2x e^{x^2} dx$.

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

Problem 02 Substitution

Find $\displaystyle \int \frac{x}{x^2+1} dx$.

Solution:
Let $u = x^2+1$, $du = 2x dx$ → $x dx = \frac{1}{2} du$
$\int \frac{x}{x^2+1} dx = \frac{1}{2} \int \frac{1}{u} du = \frac{1}{2} \ln|x^2+1| + C$

Problem 03 Integration by Parts

Find $\displaystyle \int x e^x dx$.

Solution:
$u = x$, $dv = e^x dx$ → $du = dx$, $v = e^x$
$\int x e^x dx = x e^x - \int e^x dx = x e^x - e^x + C = e^x(x-1) + C$

Problem 04 Integration by Parts

Find $\displaystyle \int \ln x dx$.

Solution:
$u = \ln x$, $dv = dx$ → $du = \frac{1}{x} dx$, $v = x$
$\int \ln x dx = x \ln x - \int x \cdot \frac{1}{x} dx = x \ln x - x + C$

Problem 05 Long Division

Find $\displaystyle \int \frac{x^2}{x+1} dx$.

Solution:
Divide: $x^2 \div (x+1) = x - 1 + \frac{1}{x+1}$
$\int \frac{x^2}{x+1} dx = \int (x - 1) dx + \int \frac{1}{x+1} dx = \frac{x^2}{2} - x + \ln|x+1| + C$

Problem 06 Partial Fractions

Find $\displaystyle \int \frac{1}{x^2 - 4} dx$.

Solution:
$\frac{1}{(x-2)(x+2)} = \frac{A}{x-2} + \frac{B}{x+2}$ → $A = \frac{1}{4}$, $B = -\frac{1}{4}$
$\int \frac{1}{x^2-4} dx = \frac{1}{4} \ln|x-2| - \frac{1}{4} \ln|x+2| + C = \frac{1}{4} \ln\left|\frac{x-2}{x+2}\right| + C$

Problem 07 Partial Fractions

Find $\displaystyle \int \frac{x+1}{(x-2)^2} dx$.

Solution:
$\frac{x+1}{(x-2)^2} = \frac{A}{x-2} + \frac{B}{(x-2)^2}$ → $A=1$, $B=3$
$\int \frac{x+1}{(x-2)^2} dx = \ln|x-2| - \frac{3}{x-2} + C$

Problem 08 Partial Fractions

Find $\displaystyle \int \frac{2x+1}{x^2+1} dx$.

Solution:
$\int \frac{2x}{x^2+1} dx + \int \frac{1}{x^2+1} dx = \ln(x^2+1) + \arctan x + C$

Problem 09 Trig Substitution

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

Solution:
$x = 3\sin\theta$, $dx = 3\cos\theta d\theta$, $\sqrt{9-x^2} = 3\cos\theta$
$\int \frac{1}{3\cos\theta} \cdot 3\cos\theta d\theta = \int d\theta = \theta + C = \arcsin\left(\frac{x}{3}\right) + C$

Problem 10 Trig Substitution

Find $\displaystyle \int \frac{1}{x^2\sqrt{x^2+4}} dx$.

Solution:
$x = 2\tan\theta$, $dx = 2\sec^2\theta d\theta$, $\sqrt{x^2+4} = 2\sec\theta$
$\int \frac{2\sec^2\theta}{4\tan^2\theta \cdot 2\sec\theta} d\theta = \frac{1}{4} \int \frac{\sec\theta}{\tan^2\theta} d\theta = \frac{1}{4} \int \frac{\cos\theta}{\sin^2\theta} d\theta = -\frac{1}{4\sin\theta} + C = -\frac{\sqrt{x^2+4}}{4x} + C$

Problem 11 Trig Substitution

Find $\displaystyle \int \frac{1}{\sqrt{x^2-16}} dx$ for $x > 4$.

Solution:
$x = 4\sec\theta$, $dx = 4\sec\theta\tan\theta d\theta$, $\sqrt{x^2-16} = 4\tan\theta$
$\int \frac{4\sec\theta\tan\theta}{4\tan\theta} d\theta = \int \sec\theta d\theta = \ln|\sec\theta + \tan\theta| + C = \ln\left|\frac{x}{4} + \frac{\sqrt{x^2-16}}{4}\right| + C = \ln\left|x + \sqrt{x^2-16}\right| + C'$

Problem 12 Improper Integral

Evaluate $\displaystyle \int_1^\infty \frac{1}{x^3} dx$ or show it diverges.

Solution:
$\int_1^b x^{-3} dx = \left[-\frac{1}{2x^2}\right]_1^b = -\frac{1}{2b^2} + \frac{1}{2}$
$\lim_{b \to \infty} \left(\frac{1}{2} - \frac{1}{2b^2}\right) = \frac{1}{2}$ → converges to $\frac{1}{2}$

Problem 13 Improper Integral

Evaluate $\displaystyle \int_1^\infty \frac{1}{x} dx$ or show it diverges.

Solution:
$\int_1^b \frac{1}{x} dx = \ln b$
$\lim_{b \to \infty} \ln b = \infty$ → diverges

Problem 14 Improper Integral

Evaluate $\displaystyle \int_0^1 \frac{1}{\sqrt{1-x}} dx$ or show it diverges.

Solution:
$\int_0^t (1-x)^{-1/2} dx = \left[-2(1-x)^{1/2}\right]_0^t = -2\sqrt{1-t} + 2$
$\lim_{t \to 1^-} (2 - 2\sqrt{1-t}) = 2$ → converges to $2$

Problem 15 Improper Integral

Evaluate $\displaystyle \int_0^1 \frac{1}{x} dx$ or show it diverges.

Solution:
$\int_t^1 \frac{1}{x} dx = -\ln t$
$\lim_{t \to 0^+} (-\ln t) = \infty$ → diverges

Problem 16 Improper Integral

Evaluate $\displaystyle \int_{-\infty}^\infty \frac{1}{1+x^2} dx$.

Solution:
$\int_0^\infty \frac{1}{1+x^2} dx = \lim_{b \to \infty} \arctan b = \frac{\pi}{2}$
By symmetry, $\int_{-\infty}^0 \frac{1}{1+x^2} dx = \frac{\pi}{2}$
Total = $\pi$

Problem 17 Integration by Parts

Find $\displaystyle \int x^2 \cos x dx$.

Solution:
$u = x^2$, $dv = \cos x dx$ → $du = 2x dx$, $v = \sin x$
$\int x^2 \cos x dx = x^2 \sin x - \int 2x \sin x dx$
For $\int x \sin x dx$: $u = x$, $dv = \sin x dx$ → $du = dx$, $v = -\cos x$
$= -x \cos x + \int \cos x dx = -x \cos x + \sin x + C$
Therefore: $\int x^2 \cos x dx = x^2 \sin x - 2(-x \cos x + \sin x) + C = x^2 \sin x + 2x \cos x - 2 \sin x + C$

Problem 18 Improper Integral

Evaluate $\displaystyle \int_0^2 \frac{1}{x-1} dx$ or show it diverges.

Solution:
Split at $x=1$: $\int_0^1 \frac{1}{x-1} dx + \int_1^2 \frac{1}{x-1} dx$
$\int_0^1 \frac{1}{x-1} dx = \lim_{t \to 1^-} [\ln|x-1|]_0^t = \lim_{t \to 1^-} (\ln(1-t) - \ln 1) = -\infty$
Since one part diverges, the whole integral diverges.

Problem 19 Partial Fractions

Find $\displaystyle \int \frac{2x+3}{x^2+2x+5} dx$.

Solution:
Complete the square: $x^2+2x+5 = (x+1)^2 + 4$
$\int \frac{2x+3}{x^2+2x+5} dx = \int \frac{2x+2}{x^2+2x+5} dx + \int \frac{1}{(x+1)^2+4} dx$
$= \ln|x^2+2x+5| + \frac{1}{2} \arctan\left(\frac{x+1}{2}\right) + C$

Problem 20 Trig Substitution

Evaluate $\displaystyle \int_0^2 \sqrt{4 - x^2} dx$.

Solution:
$x = 2\sin\theta$, $dx = 2\cos\theta d\theta$, limits: $x=0 \to \theta=0$, $x=2 \to \theta=\frac{\pi}{2}$
$\sqrt{4-x^2} = 2\cos\theta$
$\int_0^{\pi/2} (2\cos\theta)(2\cos\theta) d\theta = 4 \int_0^{\pi/2} \cos^2\theta d\theta = 4 \cdot \frac{\pi}{4} = \pi$

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