# Antiderivatives: Reverse Power Rule, +C and Initial Value Problems > Learn how to find antiderivatives: the reverse power rule, a table of basic integrals, why you add +C, and how to solve initial value problems step by step. Source: https://practicecalculus.com/blog/antiderivatives Updated: 2026-10-30 ## Key takeaways - An antiderivative of $f$ is a function $F$ with $F'(x) = f(x)$; the indefinite integral $\int f(x)\,dx = F(x) + C$ lists all of them. - Reverse power rule: $\int x^n\,dx = \frac{x^{n+1}}{n+1} + C$ for $n \ne -1$, and $\int \frac{1}{x}\,dx = \ln|x| + C$. - The $+C$ is there because constants have derivative zero, so every antiderivative can be shifted up or down. - An initial value problem gives one point on the curve, which pins down the exact value of $C$. - Always check an antiderivative by differentiating it. You can differentiate $x^3$ in your sleep. Now the question runs backwards: what function has derivative $3x^2$? That reverse question is the start of integral calculus, and the answers, called **antiderivatives**, come from running the derivative rules you already know in reverse. ## What is an antiderivative? A function $F$ is an **antiderivative** of $f$ if $$F'(x) = f(x)$$ For example, $F(x) = x^3$ is an antiderivative of $f(x) = 3x^2$, because $\frac{d}{dx}x^3 = 3x^2$. But $x^3 + 1$ and $x^3 - 8$ work too. Any constant disappears when you differentiate. So we write the **indefinite integral**, which collects every antiderivative at once: $$\int 3x^2\,dx = x^3 + C$$ The $\int$ sign and the $dx$ go together like brackets: $dx$ says the variable is $x$, and everything between them is the function you are antidifferentiating (the **integrand**). ## Why do we add +C? If $F'(x) = f(x)$, then $(F(x) + C)' = F'(x) + 0 = f(x)$ for every constant $C$. So one antiderivative gives infinitely many, all vertical shifts of each other. (The [mean value theorem](/blog/mean-value-theorem) proves that on an interval there are no others: two functions with the same derivative differ by a constant.) ```graph title: 'Antiderivatives of 2x: the parabolas y = x² + C are vertical shifts with the same slope at every x' caption: 'Every curve $y = x^2 + C$ has derivative $2x$. At $x = 1$ all of them have slope $2$, so their tangent lines are parallel.' x: [-3, 3] y: [-3, 9] functions: - expr: x^2 + 3 label: C = 3 - expr: x^2 label: C = 0 - expr: x^2 - 2 label: C = −2 points: - [1, 4, ''] - [1, 1, ''] - [1, -1, ''] ``` Leaving off $+C$ is not a small style issue. It means you only found one member of the family, and in an initial value problem it makes the answer impossible to get right. ## The reverse power rule The power rule says $\frac{d}{dx}x^{n+1} = (n+1)x^n$. Divide both sides by $n + 1$ and read it backwards: $$\int x^n\,dx = \frac{x^{n+1}}{n+1} + C \qquad (n \ne -1)$$ In words: **add one to the exponent, then divide by the new exponent.** The rule breaks at $n = -1$ because you would divide by zero. That case has its own answer: $$\int \frac{1}{x}\,dx = \ln|x| + C$$ The absolute value lets it work for negative $x$ as well. Two more rules make most problems routine. Constants factor out, and sums split: $$\int k\,f(x)\,dx = k\int f(x)\,dx, \qquad \int \big(f(x) + g(x)\big)\,dx = \int f(x)\,dx + \int g(x)\,dx$$ ## Table of basic antiderivatives Each line is a derivative rule read backwards. Check any row by differentiating the right side. | Function $f(x)$ | Antiderivative $\int f(x)\,dx$ | | ------------------------ | ------------------------------ | | $k$ (a constant) | $kx + C$ | | $x^n$, $n \ne -1$ | $\frac{x^{n+1}}{n+1} + C$ | | $\frac{1}{x}$ | $\ln\lvert x\rvert + C$ | | $e^x$ | $e^x + C$ | | $a^x$ | $\frac{a^x}{\ln a} + C$ | | $\cos x$ | $\sin x + C$ | | $\sin x$ | $-\cos x + C$ | | $\sec^2 x$ | $\tan x + C$ | | $\sec x \tan x$ | $\sec x + C$ | | $\frac{1}{1 + x^2}$ | $\arctan x + C$ | | $\frac{1}{\sqrt{1-x^2}}$ | $\arcsin x + C$ | Watch the signs on the trig rows. The derivative of $\cos x$ is $-\sin x$, so the antiderivative of $\sin x$ is $-\cos x$. ## Worked example 1: a polynomial Find $\int (3x^2 - 4x + 5)\,dx$. **Step 1. Split into terms.** Integrate $3x^2$, $-4x$ and $5$ separately. **Step 2. Reverse power rule on each.** - $\int 3x^2\,dx = 3 \cdot \frac{x^3}{3} = x^3$ - $\int -4x\,dx = -4 \cdot \frac{x^2}{2} = -2x^2$ - $\int 5\,dx = 5x$ **Step 3. Add one constant.** $$\int (3x^2 - 4x + 5)\,dx = x^3 - 2x^2 + 5x + C$$ **Check:** $\frac{d}{dx}(x^3 - 2x^2 + 5x + C) = 3x^2 - 4x + 5$. One $+C$ at the end is enough, since several constants add up to one constant. ## Worked example 2: roots and fractions Find $\int \left(\sqrt{x} + \frac{1}{x^2}\right) dx$. **Step 1. Rewrite as powers.** $\sqrt{x} = x^{1/2}$ and $\frac{1}{x^2} = x^{-2}$. **Step 2. Reverse power rule.** - $\int x^{1/2}\,dx = \frac{x^{3/2}}{3/2} = \frac{2}{3}x^{3/2}$ - $\int x^{-2}\,dx = \frac{x^{-1}}{-1} = -\frac{1}{x}$ **Step 3. Combine.** $$\int \left(\sqrt{x} + \frac{1}{x^2}\right) dx = \frac{2}{3}x^{3/2} - \frac{1}{x} + C$$ Dividing by a fraction like $\frac{3}{2}$ is the same as multiplying by $\frac{2}{3}$. That step trips up a lot of people. ## Worked example 3: simplify first Find $\int \frac{x^2 + 1}{x}\,dx$. There is no quotient rule for integrals, so **simplify before integrating**. Divide each term by $x$: $$\frac{x^2 + 1}{x} = x + \frac{1}{x}$$ Now integrate term by term: $$\int \left(x + \frac{1}{x}\right) dx = \frac{x^2}{2} + \ln|x| + C$$ ## Worked example 4: trig and exponentials Find $\int (2\cos x - 3e^x)\,dx$. Using the table, $\int \cos x\,dx = \sin x$ and $\int e^x\,dx = e^x$: $$\int (2\cos x - 3e^x)\,dx = 2\sin x - 3e^x + C$$ **Check:** $\frac{d}{dx}(2\sin x - 3e^x) = 2\cos x - 3e^x$. Two more table entries in action: find $\int \left(\frac{4}{1 + x^2} + \sec^2 x\right) dx$. Each term is a row of the table with a constant in front: $$\int \left(\frac{4}{1 + x^2} + \sec^2 x\right) dx = 4\arctan x + \tan x + C$$ ## How to check any antiderivative Integration is hard; differentiation is easy. So every antiderivative comes with a free check: differentiate your answer and see whether you get the integrand back. If $\frac{d}{dx}F(x) = f(x)$, you are right, no matter what method you used. This catches almost every slip. Suppose you wrote $\int x^{-1/2}\,dx = \frac{x^{1/2}}{1/2}$ and then simplified it to $\frac{1}{2}\sqrt{x}$. Differentiating gives $\frac{1}{4}x^{-1/2}$, which is not $x^{-1/2}$, so you know something went wrong (the correct answer is $2\sqrt{x} + C$). Make the check a habit on every problem until the rules feel automatic. ## Initial value problems: finding the exact C An **initial value problem** gives you a derivative and one point on the original function. The point picks out exactly one curve from the family $F(x) + C$. ### Worked example 5: find f from f' and one point Find $f(x)$ if $f'(x) = 6x^2 - 2x$ and $f(1) = 4$. **Step 1. Antidifferentiate.** $$f(x) = 2x^3 - x^2 + C$$ **Step 2. Use the point.** Substitute $x = 1$ and $f(1) = 4$: $$2(1)^3 - (1)^2 + C = 4 \quad\Rightarrow\quad 1 + C = 4 \quad\Rightarrow\quad C = 3$$ **Step 3. Write the answer.** $$f(x) = 2x^3 - x^2 + 3$$ ### Worked example 6: motion from acceleration A ball is thrown straight up from the ground at 48 ft/s. Its acceleration is $a(t) = -32$ ft/s². Find its height $s(t)$ and its maximum height. **Step 1. Velocity.** $v(t) = \int -32\,dt = -32t + C_1$. At launch $v(0) = 48$, so $C_1 = 48$ and $v(t) = -32t + 48$. **Step 2. Position.** $s(t) = \int (-32t + 48)\,dt = -16t^2 + 48t + C_2$. It starts on the ground, $s(0) = 0$, so $C_2 = 0$: $$s(t) = -16t^2 + 48t$$ **Step 3. Maximum height.** The ball peaks when $v(t) = 0$, at $t = 1.5$ s. Then $s(1.5) = -16(2.25) + 72 = 36$ ft. Each antiderivative needed its own constant, and each was fixed by its own initial condition. ### Worked example 7: starting from the second derivative Find $f(x)$ if $f''(x) = 12x$, $f'(0) = 1$ and $f(0) = 2$. **Step 1.** $f'(x) = 6x^2 + C_1$. Since $f'(0) = 1$, $C_1 = 1$, so $f'(x) = 6x^2 + 1$. **Step 2.** $f(x) = 2x^3 + x + C_2$. Since $f(0) = 2$, $C_2 = 2$. $$f(x) = 2x^3 + x + 2$$ **Check:** $f'(x) = 6x^2 + 1$ and $f''(x) = 12x$, with $f'(0) = 1$ and $f(0) = 2$. Two integrations always need two conditions. ## Common antiderivative mistakes 1. **Forgetting $+C$.** It costs marks on indefinite integrals and ruins initial value problems. 2. **Using the reverse power rule on $\frac{1}{x}$.** $\frac{x^0}{0}$ is undefined. The answer is $\ln|x| + C$. 3. **Integrating products or quotients piece by piece.** $\int x \cdot x^2\,dx$ is not $\frac{x^2}{2}\cdot\frac{x^3}{3}$. Multiply out first to get $\int x^3\,dx = \frac{x^4}{4} + C$. 4. **Sign slips with trig.** $\int \sin x\,dx = -\cos x + C$, not $\cos x + C$. 5. **Not rewriting roots and fractions.** $\int \frac{1}{\sqrt{x}}\,dx$ looks hard until you write it as $\int x^{-1/2}\,dx = 2\sqrt{x} + C$. When the integrand is a composite function, such as $(3x + 1)^5$, the basic table is not enough. That is the job of [u-substitution](/blog/u-substitution). And antiderivatives are exactly what you need to compute areas with the [fundamental theorem of calculus](/blog/fundamental-theorem-of-calculus). ## Practice until it is automatic Antiderivatives are the foundation of every integral you will ever compute, so speed here pays off for the rest of the course. Work through the interactive lesson below: each answer is checked instantly, missing constants are flagged, and every solution is shown step by step. ## Frequently asked questions ### What is an antiderivative? An antiderivative of $f(x)$ is any function whose derivative is $f(x)$. For example $x^3$ is an antiderivative of $3x^2$, and so is $x^3 + 7$. Finding antiderivatives is the reverse of differentiating. ### Why do you add +C to an antiderivative? Because the derivative of any constant is zero, $x^2$, $x^2 + 5$ and $x^2 - 100$ all have derivative $2x$. The $+C$ stands for every possible constant, so the answer includes all antiderivatives, not just one. ### What is the reverse power rule? To integrate $x^n$, add one to the exponent and divide by the new exponent: $\int x^n\,dx = \frac{x^{n+1}}{n+1} + C$. It works for every real $n$ except $n = -1$, where you get $\ln|x| + C$ instead. ### Is an antiderivative the same as an indefinite integral? Nearly. An antiderivative is one function, like $x^3$. The indefinite integral $\int 3x^2\,dx = x^3 + C$ is the whole family of antiderivatives. In practice the two words are used almost interchangeably. ### How do you solve an initial value problem? Find the general antiderivative with $+C$, then substitute the given point, such as $f(1) = 4$, and solve for $C$. Write the final function with that specific number in place of $C$.