Antiderivative Practice Problems with Answers and Solutions

These antiderivative practice problems start with the power rule and build up to trig, exponential and initial value problems. Try each one before you read the solution. The answers come first so you can check quickly, and the full solutions follow.

The rule you will use most:

Add one to the exponent, divide by the new exponent, and add +C. The one exception is n = -1, where \int \frac{1}{x}\,dx = \ln|x| + C. For a refresher, read the reverse power rule page.

Problems

Power rule

1. \displaystyle\int (4x^3 - 6x + 2)\,dx
2. \displaystyle\int 3\sqrt{x}\,dx
3. \displaystyle\int \frac{1}{x^3}\,dx
4. \displaystyle\int \frac{5}{x}\,dx

Trig and exponential

5. \displaystyle\int (2e^x + \cos x)\,dx
6. \displaystyle\int (\sec^2 x - 3\sin x)\,dx
7. \displaystyle\int \frac{4}{1 + x^2}\,dx

Simplify first

8. \displaystyle\int (x^2 + 1)^2\,dx
9. \displaystyle\int \frac{x^2 + 3}{\sqrt{x}}\,dx

Initial value problems

10. Find f(x) if f'(x) = 6x^2 - 4 and f(1) = 5.
11. Find f(x) if f''(x) = 12x, f'(0) = 3 and f(0) = -1.
12. A particle has velocity v(t) = 3t^2 - 2t and position s(0) = 4. Find s(2).

Answers

Step-by-step solutions

1. Integrate term by term. 4x^3 becomes x^4, -6x becomes -3x^2, and 2 becomes 2x:

2. Write \sqrt{x} = x^{1/2}. The new exponent is \frac{3}{2}, so divide by \frac{3}{2}:

3. Write \frac{1}{x^3} = x^{-3}. The new exponent is -2:

4. This is the n = -1 case, so the power rule does not apply. Use the log rule (see the integral of 1/x):

5. The derivative of e^x is e^x and the derivative of \sin x is \cos x, so:

6. The derivative of \tan x is \sec^2 x. The antiderivative of \sin x is -\cos x, so -3\sin x gives +3\cos x:

7. The derivative of \arctan x is \frac{1}{1+x^2}:

8. Expand first: (x^2+1)^2 = x^4 + 2x^2 + 1. Then integrate each term:

9. Split the fraction: \frac{x^2}{x^{1/2}} + \frac{3}{x^{1/2}} = x^{3/2} + 3x^{-1/2}. Then:

10. The general antiderivative is f(x) = 2x^3 - 4x + C. Use f(1) = 5: 2 - 4 + C = 5, so C = 7:

11. Integrate twice, using one condition each time. First f'(x) = 6x^2 + C_1, and f'(0) = 3 gives C_1 = 3. Then f(x) = 2x^3 + 3x + C_2, and f(0) = -1 gives C_2 = -1:

12. Position is an antiderivative of velocity: s(t) = t^3 - t^2 + C. Since s(0) = 4, C = 4. Then:

Common mistakes on these problems

• Forgetting +C. An indefinite integral is a whole family of functions. Leaving off +C loses marks on most tests.
• Using the power rule on \frac{1}{x}. It would give \frac{x^0}{0}, which is undefined. The answer is \ln|x| + C.
• Sign errors with sine and cosine. The antiderivative of \sin x is -\cos x, not \cos x. Differentiate your answer to check.
• Integrating a product factor by factor. In problem 8, \int (x^2+1)^2\,dx is not \frac{(x^2+1)^3}{3}. Expand first, or check by differentiating.
• Solving for C too late. In problem 11, find C_1 before you integrate again, or you will end up with an unknown C_1 x term.

Keep practicing

These 12 problems use fixed numbers. For unlimited practice with new numbers and instant feedback, try the reverse power rule lesson and the trig and exponential integrals lesson. When antiderivatives feel easy, move on to definite integral problems and u-substitution problems. For the ideas behind it, read the antiderivatives guide.