# Antiderivative Practice Problems with Answers and Solutions > 12 antiderivative practice problems with answers and step-by-step solutions: power rule, trig, exponentials and initial value problems. Source: https://practicecalculus.com/practice-problems/antiderivatives Updated: 2026-11-06 $$ \int x^n\,dx = \frac{x^{n+1}}{n+1} + C \quad (n \neq -1) $$ ## At a glance - **Problems:** 12, with answers and full solutions - **Level:** Calculus 1 and AP Calculus AB, Unit 6 - **You need:** The [reverse power rule](/rules/reverse-power-rule) and the basic derivatives - **Main idea:** Ask which function has this derivative, then add $+C$ - **Practice with new numbers:** [Antiderivative lessons](/learn/calculus-1/integrals/undoing-derivatives) ## Key takeaways - Use the reverse power rule term by term: add one to the exponent and divide by the new exponent. - Rewrite roots and fractions as powers before you integrate, and remember that $\int \frac{1}{x}\,dx = \ln|x| + C$. - In an initial value problem, find the general antiderivative first, then use the given point to solve for $C$. 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: $$\int x^n\,dx = \frac{x^{n+1}}{n+1} + C \qquad (n \neq -1)$$ 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](/rules/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 | Problem | Answer | | --- | --- | | 1 | $x^4 - 3x^2 + 2x + C$ | | 2 | $2x^{3/2} + C$ | | 3 | $-\dfrac{1}{2x^2} + C$ | | 4 | $5\ln\lvert x\rvert + C$ | | 5 | $2e^x + \sin x + C$ | | 6 | $\tan x + 3\cos x + C$ | | 7 | $4\arctan x + C$ | | 8 | $\dfrac{x^5}{5} + \dfrac{2x^3}{3} + x + C$ | | 9 | $\dfrac{2}{5}x^{5/2} + 6x^{1/2} + C$ | | 10 | $f(x) = 2x^3 - 4x + 7$ | | 11 | $f(x) = 2x^3 + 3x - 1$ | | 12 | $s(2) = 8$ | ## Step-by-step solutions **1.** Integrate term by term. $4x^3$ becomes $x^4$, $-6x$ becomes $-3x^2$, and $2$ becomes $2x$: $$\int (4x^3 - 6x + 2)\,dx = x^4 - 3x^2 + 2x + C$$ **2.** Write $\sqrt{x} = x^{1/2}$. The new exponent is $\frac{3}{2}$, so divide by $\frac{3}{2}$: $$3\cdot\frac{x^{3/2}}{3/2} + C = 2x^{3/2} + C$$ **3.** Write $\frac{1}{x^3} = x^{-3}$. The new exponent is $-2$: $$\frac{x^{-2}}{-2} + C = -\frac{1}{2x^2} + C$$ **4.** This is the $n = -1$ case, so the power rule does not apply. Use the log rule (see the [integral of 1/x](/integrals/1-over-x)): $$\int \frac{5}{x}\,dx = 5\ln|x| + C$$ **5.** The derivative of $e^x$ is $e^x$ and the derivative of $\sin x$ is $\cos x$, so: $$\int (2e^x + \cos x)\,dx = 2e^x + \sin x + C$$ **6.** The derivative of $\tan x$ is $\sec^2 x$. The antiderivative of $\sin x$ is $-\cos x$, so $-3\sin x$ gives $+3\cos x$: $$\int (\sec^2 x - 3\sin x)\,dx = \tan x + 3\cos x + C$$ **7.** The derivative of $\arctan x$ is $\frac{1}{1+x^2}$: $$\int \frac{4}{1+x^2}\,dx = 4\arctan x + C$$ **8.** Expand first: $(x^2+1)^2 = x^4 + 2x^2 + 1$. Then integrate each term: $$\frac{x^5}{5} + \frac{2x^3}{3} + x + C$$ **9.** Split the fraction: $\frac{x^2}{x^{1/2}} + \frac{3}{x^{1/2}} = x^{3/2} + 3x^{-1/2}$. Then: $$\frac{x^{5/2}}{5/2} + 3\cdot\frac{x^{1/2}}{1/2} + C = \frac{2}{5}x^{5/2} + 6x^{1/2} + C$$ **10.** The general antiderivative is $f(x) = 2x^3 - 4x + C$. Use $f(1) = 5$: $2 - 4 + C = 5$, so $C = 7$: $$f(x) = 2x^3 - 4x + 7$$ **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 $C_2 = -1$: $$f(x) = 2x^3 + 3x - 1$$ **12.** Position is an antiderivative of velocity: $s(t) = t^3 - t^2 + C$. Since $s(0) = 4$, $C = 4$. Then: $$s(2) = 8 - 4 + 4 = 8$$ ## 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](/learn/calculus-1/integrals/reverse-power-rule) and the [trig and exponential integrals lesson](/learn/calculus-1/integrals/trig-exponential-integrals). When antiderivatives feel easy, move on to [definite integral problems](/practice-problems/definite-integrals) and [u-substitution problems](/practice-problems/u-substitution). For the ideas behind it, read the [antiderivatives guide](/blog/antiderivatives). ## Frequently asked questions ### How do you check an antiderivative? Differentiate your answer. If you get back the function you started with, the antiderivative is right. This check takes a few seconds and catches most sign and coefficient errors. ### Why do you add +C to an antiderivative? Every constant has derivative zero, so $x^2$, $x^2 + 5$ and $x^2 - 3$ all have derivative $2x$. The $+C$ stands for all of them at once. An initial condition picks out one value of $C$. ### Is there a product rule for antiderivatives? No. The integral of a product is not the product of the integrals. Expand or simplify first, as in problem 8 below, or use [u-substitution](/practice-problems/u-substitution) or integration by parts for harder products.