AP Calculus AB Integration by Substitution Worksheet Solutions (PDF)

AP Calculus AB math worksheet solutions: Integration by Substitution – Maths PDF

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This u substitution answers for AP Calculus AB covers integration by substitution (Common Core AP CALC AB 6.9, AP CALC AB 6.10, AP CALC AB 6.11, AP CALC AB 6.12, AP CALC AB 6.14), with a free PDF to print.

This answer key gives a full worked solution for every problem in the u-substitution worksheet. Compare each step, not only the final answer, because AP graders award credit for a correct setup and clear justification. Final answers are in bold. Redo any problem you missed before moving on to the test.

AP Calculus AB · Common Core State Standards: AP CALC AB 6.9, AP CALC AB 6.10, AP CALC AB 6.11, AP CALC AB 6.12, AP CALC AB 6.14

U substitution answers: exercises 1 to 10

Exercise 1 – A power of a quadratic

Let \(u=x^{2}+1\). Then \(du=2x\,dx\).

The integral becomes \(\int u^{3}du=\frac{u^{4}}{4}\).

Back-substitute \(u=x^{2}+1\) and add the constant.

Final answer: \(\frac{(x^{2}+1)^{4}}{4}+C\).

Exercise 2 – A square root

Let \(u=x^{2}+4\). Then \(du=2x\,dx\).

The integral becomes \(\frac{1}{2}\int u^{1/2}du=\frac{1}{2}\cdot\frac{2}{3}u^{3/2}\).

Back-substitute \(u=x^{2}+4\) and add the constant.

Final answer: \(\frac{1}{3}(x^{2}+4)^{3/2}+C\).

Exercise 3 – Cosine of a multiple

Let \(u=3x\). Then \(du=3\,dx\).

The integral becomes \(\frac{1}{3}\int\cos u\,du=\frac{1}{3}\sin u\).

Back-substitute \(u=3x\) and add the constant.

Final answer: \(\frac{1}{3}\sin 3x+C\).

cos 3x (blue) and the antiderivative (sin 3x)/3 (green).

Exercise 4 – Exponential of a multiple

Let \(u=5x\). Then \(du=5\,dx\).

The integral becomes \(\frac{1}{5}\int e^{u}du=\frac{1}{5}e^{u}\).

Back-substitute \(u=5x\) and add the constant.

Final answer: \(\frac{1}{5}e^{5x}+C\).

Exercise 5 – Exponential with a quadratic

Let \(u=x^{2}\). Then \(du=2x\,dx\).

The integral becomes \(\frac{1}{2}\int e^{u}du=\frac{1}{2}e^{u}\).

Back-substitute \(u=x^{2}\) and add the constant.

Final answer: \(\frac{1}{2}e^{x^{2}}+C\).

Area: \(\left[\frac{1}{2}e^{x^{2}}\right]_{0}^{1}=\frac{e-1}{2}\approx 0.859\).

Exercise 6 – Sine times a power of cosine

Let \(u=\cos x\). Then \(du=-\sin x\,dx\).

The integral becomes \(-\int u^{2}du=-\frac{u^{3}}{3}\).

Back-substitute \(u=\cos x\) and add the constant.

Final answer: \(-\frac{\cos^{3}x}{3}+C\).

Exercise 7 – Logarithm over x

Let \(u=\ln x\). Then \(du=\frac{1}{x}\,dx\).

The integral becomes \(\int u\,du=\frac{u^{2}}{2}\).

Back-substitute \(u=\ln x\) and add the constant.

Final answer: \(\frac{(\ln x)^{2}}{2}+C\).

Exercise 8 – Reciprocal of a linear function

Let \(u=2x+3\). Then \(du=2\,dx\).

The integral becomes \(\frac{1}{2}\int\frac{1}{u}du=\frac{1}{2}\ln|u|\).

Back-substitute \(u=2x+3\) and add the constant.

Final answer: \(\frac{1}{2}\ln|2x+3|+C\).

Exercise 9 – Quotient with a quadratic

Let \(u=x^{2}+1\). Then \(du=2x\,dx\).

The integral becomes \(\frac{1}{2}\int\frac{1}{u}du=\frac{1}{2}\ln|u|\).

Back-substitute \(u=x^{2}+1\) and add the constant.

Final answer: \(\frac{1}{2}\ln(x^{2}+1)+C\).

Exercise 10 – Integral of tangent

Write \(\tan x=\frac{\sin x}{\cos x}\). Let \(u=\cos x\), \(du=-\sin x\,dx\).

\(-\int\frac{1}{u}du=-\ln|u|\).

Final answer: \(-\ln|\cos x|+C\).

tan x (blue) and an antiderivative −ln|cos x| (green).

Exercises 11 and up

Exercise 11 – Exponential over one plus exponential

Let \(u=1+e^{x}\). Then \(du=e^{x}\,dx\).

The integral becomes \(\int\frac{1}{u}du=\ln|u|\).

Back-substitute \(u=1+e^{x}\) and add the constant.

Final answer: \(\ln(1+e^{x})+C\).

Exercise 12 – Secant squared of a multiple

Let \(u=2x\). Then \(du=2\,dx\).

The integral becomes \(\frac{1}{2}\int\sec^{2}u\,du=\frac{1}{2}\tan u\).

Back-substitute \(u=2x\) and add the constant.

Final answer: \(\frac{1}{2}\tan 2x+C\).

Exercise 13 – Inverse tangent form

Let \(u=3x\). Then \(du=3\,dx\).

The integral becomes \(\frac{1}{3}\int\frac{1}{1+u^{2}}du=\frac{1}{3}\tan^{-1}u\).

Back-substitute \(u=3x\) and add the constant.

Final answer: \(\frac{1}{3}\tan^{-1}(3x)+C\).

Exercise 14 – Cubic inside a power

Let \(u=x^{3}+1\). Then \(du=3x^{2}\,dx\).

The integral becomes \(\frac{1}{3}\int u^{4}du=\frac{1}{3}\cdot\frac{u^{5}}{5}\).

Back-substitute \(u=x^{3}+1\) and add the constant.

Final answer: \(\frac{(x^{3}+1)^{5}}{15}+C\).

Exercise 15 – Definite integral with changed limits

Let \(u=x^{2}+1\), \(du=2x\,dx\), so \(x\,dx=\frac{1}{2}du\). When \(x=0\), \(u=1\); when \(x=2\), \(u=5\).

\(\frac{1}{2}\int_{1}^{5}u^{3}du=\frac{1}{2}\left[\frac{u^{4}}{4}\right]_{1}^{5}=\frac{625-1}{8}=78\).

Final answer: \(78\).

Exercise 16 – Definite integral, trig

Let \(u=\sin x\), \(du=\cos x\,dx\). Limits: \(0\to 0\) and \(\frac{\pi}{2}\to 1\).

\(\int_{0}^{1}u\,du=\frac{1}{2}\).

Final answer: \(\frac{1}{2}\).

Exercise 17 – Definite integral, logarithm

Let \(u=\ln x\), \(du=\frac{1}{x}dx\). Limits: \(1\to 0\) and \(e\to 1\).

\(\int_{0}^{1}u\,du=\frac{1}{2}\).

Final answer: \(\frac{1}{2}\).

Exercise 18 – Definite integral, rational

Let \(u=x^{3}+2\), \(du=3x^{2}dx\). Limits: \(-1\to 1\) and \(1\to 3\).

\(\frac{1}{3}\int_{1}^{3}\frac{1}{u}du=\frac{1}{3}(\ln 3-\ln 1)=\frac{\ln 3}{3}\).

Final answer: \(\frac{\ln 3}{3}\approx 0.366\).

Exercise 19 – Long division

Divide: \(x^{2}+3x+1=(x+1)(x+2)-1\), so \(\frac{x^{2}+3x+1}{x+1}=x+2-\frac{1}{x+1}\).

Integrate term by term.

Final answer: \(\frac{x^{2}}{2}+2x-\ln|x+1|+C\).

The quotient x + 2 is the slant line the graph approaches.

Exercise 20 – Completing the square

Complete the square: \(x^{2}+4x+13=(x+2)^{2}+9\).

Let \(u=x+2\): \(\int\frac{1}{u^{2}+9}du=\frac{1}{3}\tan^{-1}\frac{u}{3}\).

Final answer: \(\frac{1}{3}\tan^{-1}\left(\frac{x+2}{3}\right)+C\).

Exercise 21 – Colony growth

Let \(u=-0.1t^{2}\), \(du=-0.2t\,dt\), so \(t\,dt=-5\,du\). An antiderivative is \(-1000e^{-0.1t^{2}}\).

\(\left[-1000e^{-0.1t^{2}}\right]_{0}^{3}=-1000e^{-0.9}+1000=1000(1-e^{-0.9})\approx 593.4\).

Final answer: \(\approx 593.4\text{ bacteria}\).

Exercise 22 – Battery charge

Let \(u=-0.5t\), \(dt=-2\,du\). An antiderivative is \(-10e^{-0.5t}\).

\(\left[-10e^{-0.5t}\right]_{0}^{4}=-10e^{-2}+10=10(1-e^{-2})\approx 8.65\).

Final answer: \(\approx 8.65\text{ amp-hours}\).

Exercise 23 – Distance from a velocity

Let \(u=t^{2}+16\), \(du=2t\,dt\), so \(3t\,dt=\frac{3}{2}du\). Limits: \(0\to 16\) and \(3\to 25\).

\(\frac{3}{2}\int_{16}^{25}u^{1/2}du=\frac{3}{2}\cdot\frac{2}{3}\left[u^{3/2}\right]_{16}^{25}=125-64=61\).

Final answer: \(61\text{ feet}\).

Key facts behind this u substitution answers

Key facts

Let u = g(x) and du = g′(x) dx, then the integral becomes the integral of f(u) du.
Adjust constant factors: if du is off by a constant, divide or multiply to compensate.
For definite integrals, change the limits to u-values or return to x before evaluating.
Use long division for improper rational functions and completing the square for irreducible quadratics.

How to use this u substitution answers

Cover each solution, solve the exercise again, then uncover it and compare every line. If your answer differs, find the first step that changed and redo the exercise from there.

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