Worksheet · Calculus II · Unit 3A

Integration by Parts Worksheet with Complete Solutions

Twenty problems covering polynomial products, logarithms, inverse trigonometric functions, definite integrals, and cyclic cases.

20 problems70 min estimated timeIntermediate to advanced progression

What is included

Practice choosing u and dv, repeated integration by parts, and cyclic integrals with a printable key.

Skills assessed

LIATE judgment
repeated integration by parts
definite integral evaluation

Prerequisites

antiderivatives
product rule

Integration by Parts Worksheet instructional sequence — Twenty problems covering polynomial products, logarithms, inverse trigonometric functions, definite integrals, and cyclic cases. The numbered labels and written sequence preserve meaning without relying on color.

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Evaluate $\int xe^x\,dx$ .
Evaluate $\int x^2e^x\,dx$ .
Evaluate $\int x^3e^x\,dx$ .
Evaluate $\int xe^{2x}\,dx$ .
Evaluate $\int x\sin x\,dx$ .
Evaluate $\int x\cos x\,dx$ .
Evaluate $\int x^2\sin x\,dx$ .
Evaluate $\int x^2\cos x\,dx$ .
Evaluate $\int\ln x\,dx$ .
Evaluate $\int x\ln x\,dx$ .
Evaluate $\int(\ln x)^2\,dx$ .
Evaluate $\int\arctan x\,dx$ .
Evaluate $\int\arcsin x\,dx$ .
Evaluate $\int_0^1xe^x\,dx$ .
Evaluate $\int_0^\pi x\sin x\,dx$ .
Evaluate $\int e^x\cos x\,dx$ .
Evaluate $\int e^x\sin x\,dx$ .
Evaluate $\int x^4e^x\,dx$ .
Evaluate $\int x^2\ln x\,dx$ .
Evaluate $\int_1^e\ln x\,dx$ .

Complete worked solutions

Every problem has a source-matched answer and independently reviewed derivation.

Problem 1: Evaluate $\int xe^x\,dx$ .

Answer: $e^x(x-1)+C$

Why this method: Integration by parts matches the mathematical structure before any algebraic cleanup.

Take $u=x$ and $dv=e^x dx$ , so $du=dx$ and $v=e^x$ .
Then $\int xe^x dx=xe^x-\int e^x dx=xe^x-e^x+C=e^x(x-1)+C$ .
Therefore the result is $e^x(x-1)+C$ .

Problem 2: Evaluate $\int x^2e^x\,dx$ .

Answer: $e^x(x^2-2x+2)+C$

Why this method: Integration by parts matches the mathematical structure before any algebraic cleanup.

Take $u=x^2$ and $dv=e^x dx$ : $I=x^2e^x-2\int xe^x dx$ .
Using $\int xe^x dx=e^x(x-1)$ , $I=e^x(x^2-2x+2)+C$ .
Therefore the result is $e^x(x^2-2x+2)+C$ .

Problem 3: Evaluate $\int x^3e^x\,dx$ .

Answer: $e^x(x^3-3x^2+6x-6)+C$

Why this method: Integration by parts matches the mathematical structure before any algebraic cleanup.

Take $u=x^3$ and $dv=e^x dx$ : $I=x^3e^x-3\int x^2e^x dx$ .
Substitute $\int x^2e^x dx=e^x(x^2-2x+2)$ and collect terms to get $e^x(x^3-3x^2+6x-6)+C$ .

Problem 4: Evaluate $\int xe^{2x}\,dx$ .

Answer: $e^{2x}(2x-1)/4+C$

Why this method: Integration by parts matches the mathematical structure before any algebraic cleanup.

Take $u=x$ and $dv=e^{2x}dx$ , so $du=dx$ and $v=e^{2x}/2$ .
Thus $I=xe^{2x}/2-\int e^{2x}/2\,dx=xe^{2x}/2-e^{2x}/4+C=e^{2x}(2x-1)/4+C$ .
Therefore the result is $e^{2x}(2x-1)/4+C$ .

Problem 5: Evaluate $\int x\sin x\,dx$ .

Answer: $-x\cos x+\sin x+C$

Why this method: Integration by parts matches the mathematical structure before any algebraic cleanup.

Take $u=x$ and $dv=\sin x\,dx$ , so $du=dx$ and $v=-\cos x$ .
Then $I=-x\cos x+\int\cos x\,dx=-x\cos x+\sin x+C$ .
Therefore the result is $-x\cos x+\sin x+C$ .

Problem 6: Evaluate $\int x\cos x\,dx$ .

Answer: $x\sin x+\cos x+C$

Why this method: Integration by parts matches the mathematical structure before any algebraic cleanup.

Take $u=x$ and $dv=\cos x\,dx$ , so $du=dx$ and $v=\sin x$ .
Then $I=x\sin x-\int\sin x\,dx=x\sin x+\cos x+C$ .
Therefore the result is $x\sin x+\cos x+C$ .

Problem 7: Evaluate $\int x^2\sin x\,dx$ .

Answer: $-x^2\cos x+2x\sin x+2\cos x+C$

Why this method: Integration by parts matches the mathematical structure before any algebraic cleanup.

Take $u=x^2$ and $dv=\sin x\,dx$ : $I=-x^2\cos x+2\int x\cos x\,dx$ .
Since $\int x\cos x\,dx=x\sin x+\cos x$ , $I=-x^2\cos x+2x\sin x+2\cos x+C$ .
Therefore the result is $-x^2\cos x+2x\sin x+2\cos x+C$ .

Problem 8: Evaluate $\int x^2\cos x\,dx$ .

Answer: $x^2\sin x+2x\cos x-2\sin x+C$

Why this method: Integration by parts matches the mathematical structure before any algebraic cleanup.

Take $u=x^2$ and $dv=\cos x\,dx$ : $I=x^2\sin x-2\int x\sin x\,dx$ .
Since $\int x\sin x\,dx=-x\cos x+\sin x$ , $I=x^2\sin x+2x\cos x-2\sin x+C$ .
Therefore the result is $x^2\sin x+2x\cos x-2\sin x+C$ .

Problem 9: Evaluate $\int\ln x\,dx$ .

Answer: $x\ln x-x+C$

Why this method: Integration by parts matches the mathematical structure before any algebraic cleanup.

Write the integrand as $(\ln x)(1)$ ; take $u=\ln x$ and $dv=dx$ , so $du=dx/x$ and $v=x$ .
Then $I=x\ln x-\int1\,dx=x\ln x-x+C$ , for $x>0$ .
Therefore the result is $x\ln x-x+C$ .

Problem 10: Evaluate $\int x\ln x\,dx$ .

Answer: $\frac{x^2}{2}\ln x-\frac{x^2}{4}+C$

Why this method: Integration by parts matches the mathematical structure before any algebraic cleanup.

Take $u=\ln x$ and $dv=x\,dx$ , so $du=dx/x$ and $v=x^2/2$ .
Then $I=(x^2/2)\ln x-\frac12\int x\,dx=(x^2/2)\ln x-x^2/4+C$ , for $x>0$ .
Therefore the result is $\frac{x^2}{2}\ln x-\frac{x^2}{4}+C$ .

Problem 11: Evaluate $\int(\ln x)^2\,dx$ .

Answer: $x[(\ln x)^2-2\ln x+2]+C$

Why this method: Integration by parts matches the mathematical structure before any algebraic cleanup.

Take $u=(\ln x)^2$ and $dv=dx$ : $I=x(\ln x)^2-2\int\ln x\,dx$ .
Substitute $\int\ln x\,dx=x\ln x-x$ to obtain $x[(\ln x)^2-2\ln x+2]+C$ , for $x>0$ .

Problem 12: Evaluate $\int\arctan x\,dx$ .

Answer: $x\arctan x-\frac12\ln(1+x^2)+C$

Why this method: Integration by parts matches the mathematical structure before any algebraic cleanup.

Take $u=\arctan x$ and $dv=dx$ , so $du=dx/(1+x^2)$ and $v=x$ .
Then $I=x\arctan x-\int x/(1+x^2)\,dx=x\arctan x-\frac12\ln(1+x^2)+C$ .
Therefore the result is $x\arctan x-\frac12\ln(1+x^2)+C$ .

Problem 13: Evaluate $\int\arcsin x\,dx$ .

Answer: $x\arcsin x+\sqrt{1-x^2}+C$

Why this method: Integration by parts matches the mathematical structure before any algebraic cleanup.

Take $u=\arcsin x$ and $dv=dx$ , so $du=dx/\sqrt{1-x^2}$ and $v=x$ .
Because $\int x/\sqrt{1-x^2}\,dx=-\sqrt{1-x^2}$ , $I=x\arcsin x+\sqrt{1-x^2}+C$ on $(-1,1)$ .
Therefore the result is $x\arcsin x+\sqrt{1-x^2}+C$ .

Problem 14: Evaluate $\int_0^1xe^x\,dx$ .

Answer: $1$

Why this method: Integration by parts matches the mathematical structure before any algebraic cleanup.

Using $u=x$ and $dv=e^x dx$ , an antiderivative is $e^x(x-1)$ .
Evaluate the bounds: $[e^x(x-1)]_0^1=0-(-1)=1$ .
Therefore the result is $1$ .

Problem 15: Evaluate $\int_0^\pi x\sin x\,dx$ .

Answer: $\pi$

Why this method: Integration by parts matches the mathematical structure before any algebraic cleanup.

Using $u=x$ and $dv=\sin x\,dx$ , an antiderivative is $-x\cos x+\sin x$ .
Evaluate the bounds: $[-x\cos x+\sin x]_0^\pi=\pi-0=\pi$ .
Therefore the result is $\pi$ .

Problem 16: Evaluate $\int e^x\cos x\,dx$ .

Answer: $\frac{e^x}{2}(\sin x+\cos x)+C$

Why this method: Cyclic integration by parts matches the mathematical structure before any algebraic cleanup.

Let $I=\int e^x\cos x\,dx$ . Two integrations by parts give $I=e^x\cos x+e^x\sin x-I$ .
Therefore $2I=e^x(\sin x+\cos x)$ , so $I=\frac{e^x}{2}(\sin x+\cos x)+C$ .
Therefore the result is $\frac{e^x}{2}(\sin x+\cos x)+C$ .

Problem 17: Evaluate $\int e^x\sin x\,dx$ .

Answer: $\frac{e^x}{2}(\sin x-\cos x)+C$

Why this method: Cyclic integration by parts matches the mathematical structure before any algebraic cleanup.

Let $I=\int e^x\sin x\,dx$ . Two integrations by parts give $I=e^x\sin x-e^x\cos x-I$ .
Therefore $2I=e^x(\sin x-\cos x)$ , so $I=\frac{e^x}{2}(\sin x-\cos x)+C$ .
Therefore the result is $\frac{e^x}{2}(\sin x-\cos x)+C$ .

Problem 18: Evaluate $\int x^4e^x\,dx$ .

Answer: $e^x(x^4-4x^3+12x^2-24x+24)+C$

Why this method: Integration by parts matches the mathematical structure before any algebraic cleanup.

Repeatedly take the polynomial as $u$ : $I=x^4e^x-4\int x^3e^x dx$ .
Substituting the three prior reductions gives $I=e^x(x^4-4x^3+12x^2-24x+24)+C$ .
Therefore the result is $e^x(x^4-4x^3+12x^2-24x+24)+C$ .

Problem 19: Evaluate $\int x^2\ln x\,dx$ .

Answer: $\frac{x^3}{3}\ln x-\frac{x^3}{9}+C$

Why this method: Integration by parts matches the mathematical structure before any algebraic cleanup.

Take $u=\ln x$ and $dv=x^2dx$ , so $du=dx/x$ and $v=x^3/3$ .
Then $I=(x^3/3)\ln x-\frac13\int x^2dx=(x^3/3)\ln x-x^3/9+C$ , for $x>0$ .
Therefore the result is $\frac{x^3}{3}\ln x-\frac{x^3}{9}+C$ .

Problem 20: Evaluate $\int_1^e\ln x\,dx$ .

Answer: $1$

Why this method: Integration by parts matches the mathematical structure before any algebraic cleanup.

Using $u=\ln x$ and $dv=dx$ , an antiderivative is $x\ln x-x$ .
Evaluate the bounds: $[x\ln x-x]_1^e=0-(-1)=1$ .
Therefore the result is $1$ .

Common errors

Choosing dv that is not readily integrable.
Dropping the subtraction in the formula.
Failing to solve for the original integral in a cyclic case.

What is included

Skills assessed

Prerequisites

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Complete worked solutions

Problem 1: Evaluate ∫xex dx\int xe^x\,dx∫xexdx.

Problem 2: Evaluate ∫x2ex dx\int x^2e^x\,dx∫x2exdx.

Problem 3: Evaluate ∫x3ex dx\int x^3e^x\,dx∫x3exdx.

Problem 4: Evaluate ∫xe2x dx\int xe^{2x}\,dx∫xe2xdx.

Problem 5: Evaluate ∫xsin⁡x dx\int x\sin x\,dx∫xsinxdx.

Problem 6: Evaluate ∫xcos⁡x dx\int x\cos x\,dx∫xcosxdx.

Problem 7: Evaluate ∫x2sin⁡x dx\int x^2\sin x\,dx∫x2sinxdx.

Problem 8: Evaluate ∫x2cos⁡x dx\int x^2\cos x\,dx∫x2cosxdx.

Problem 9: Evaluate ∫ln⁡x dx\int\ln x\,dx∫lnxdx.

Problem 10: Evaluate ∫xln⁡x dx\int x\ln x\,dx∫xlnxdx.

Problem 11: Evaluate ∫(ln⁡x)2 dx\int(\ln x)^2\,dx∫(lnx)2dx.

Problem 12: Evaluate ∫arctan⁡x dx\int\arctan x\,dx∫arctanxdx.

Problem 13: Evaluate ∫arcsin⁡x dx\int\arcsin x\,dx∫arcsinxdx.

Problem 14: Evaluate ∫01xex dx\int_0^1xe^x\,dx∫01​xexdx.

Problem 15: Evaluate ∫0πxsin⁡x dx\int_0^\pi x\sin x\,dx∫0π​xsinxdx.

Problem 16: Evaluate ∫excos⁡x dx\int e^x\cos x\,dx∫excosxdx.

Problem 17: Evaluate ∫exsin⁡x dx\int e^x\sin x\,dx∫exsinxdx.

Problem 18: Evaluate ∫x4ex dx\int x^4e^x\,dx∫x4exdx.

Problem 19: Evaluate ∫x2ln⁡x dx\int x^2\ln x\,dx∫x2lnxdx.

Problem 20: Evaluate ∫1eln⁡x dx\int_1^e\ln x\,dx∫1e​lnxdx.

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Supporting steps

Problem 1: Evaluate $\int xe^x\,dx$ .

Problem 2: Evaluate $\int x^2e^x\,dx$ .

Problem 3: Evaluate $\int x^3e^x\,dx$ .

Problem 4: Evaluate $\int xe^{2x}\,dx$ .

Problem 5: Evaluate $\int x\sin x\,dx$ .

Problem 6: Evaluate $\int x\cos x\,dx$ .

Problem 7: Evaluate $\int x^2\sin x\,dx$ .

Problem 8: Evaluate $\int x^2\cos x\,dx$ .

Problem 9: Evaluate $\int\ln x\,dx$ .

Problem 10: Evaluate $\int x\ln x\,dx$ .

Problem 11: Evaluate $\int(\ln x)^2\,dx$ .

Problem 12: Evaluate $\int\arctan x\,dx$ .

Problem 13: Evaluate $\int\arcsin x\,dx$ .

Problem 14: Evaluate $\int_0^1xe^x\,dx$ .

Problem 15: Evaluate $\int_0^\pi x\sin x\,dx$ .

Problem 16: Evaluate $\int e^x\cos x\,dx$ .

Problem 17: Evaluate $\int e^x\sin x\,dx$ .

Problem 18: Evaluate $\int x^4e^x\,dx$ .

Problem 19: Evaluate $\int x^2\ln x\,dx$ .

Problem 20: Evaluate $\int_1^e\ln x\,dx$ .