Antiderivatives and Indefinite Integrals

An antiderivative of $f$ is a function whose derivative is $f$. Finding one is differentiation run backwards: instead of asking “what is the slope of this function?”, you ask “what function has this slope?” Antiderivatives are what make the Fundamental Theorem of Calculus work, so you need these rules at your fingertips.

Key ideas

Antiderivatives and the + C

$F$ is an antiderivative of $f$ if $F'(x) = f(x)$. For example, $x^2$ is an antiderivative of $2x$. But so are $x^2 + 5$ and $x^2 - 3$, because the derivative of a constant is $0$.

All antiderivatives of $f$ (on an interval) differ by a constant. The indefinite integral collects them all:

$$\int f(x)\,dx = F(x) + C,$$

where $C$ is any constant. The graphs of $F(x) + C$ are vertical shifts of each other.

Every curve $y = x^2 + C$ has slope $2x$. An initial condition, like passing through $(1, 3)$, picks out one of them.

Basic rules

Each rule comes from reading a derivative rule backwards. (Trig functions use radians.)

Integral	Result
$\displaystyle\int k\,dx$	$kx + C$
$\displaystyle\int x^n\,dx$, for $n \ne -1$	$\dfrac{x^{n+1}}{n+1} + C$
$\displaystyle\int \frac{1}{x}\,dx$	$\ln\lvert x\rvert + C$
$\displaystyle\int e^x\,dx$	$e^x + C$
$\displaystyle\int \sin x\,dx$	$-\cos x + C$
$\displaystyle\int \cos x\,dx$	$\sin x + C$
$\displaystyle\int \sec^2 x\,dx$	$\tan x + C$
$\displaystyle\int \sec x \tan x\,dx$	$\sec x + C$
$\displaystyle\int \csc^2 x\,dx$	$-\cot x + C$
$\displaystyle\int \csc x \cot x\,dx$	$-\csc x + C$
$\displaystyle\int \frac{1}{\sqrt{1 - x^2}}\,dx$	$\arcsin x + C$
$\displaystyle\int \frac{1}{1 + x^2}\,dx$	$\arctan x + C$

Power rule in words: add $1$ to the exponent, then divide by the new exponent. It fails for $n = -1$ (you’d divide by $0$), which is why $\tfrac{1}{x}$ has its own rule. The absolute value in $\ln|x|$ lets the answer work for negative $x$ too.

Constant multiples and sums

$$\int k\,f(x)\,dx = k\int f(x)\,dx, \qquad \int \big(f(x) \pm g(x)\big)\,dx = \int f(x)\,dx \pm \int g(x)\,dx$$

There is no product or quotient rule for integrals. Rewrite first: expand products, split fractions over a single-term denominator, and write roots as powers.

Check by differentiating

You can always check an antiderivative: differentiate your answer and see if you get the integrand back.

Initial-value problems

An initial condition such as $y(1) = 3$ picks one curve from the family:

1. Find the general antiderivative, with $+ C$.
2. Substitute the initial condition and solve for $C$.
3. Write the particular solution.

If you’re given $f''$, integrate twice, using one condition for each constant.

Worked examples

Example 1: A polynomial

Find $\displaystyle\int (6x^2 - 4x + 5)\,dx$.

Solution. Apply the power rule term by term:

$$\int (6x^2 - 4x + 5)\,dx = 6 \cdot \frac{x^3}{3} - 4 \cdot \frac{x^2}{2} + 5x + C = 2x^3 - 2x^2 + 5x + C$$

Check: $\dfrac{d}{dx}(2x^3 - 2x^2 + 5x) = 6x^2 - 4x + 5$. ✓

Example 2: Rewrite first

Find (a) $\displaystyle\int \left(\frac{3}{x^2} + \sqrt{x} - \frac{4}{x}\right)dx$ and (b) $\displaystyle\int \frac{x^2 + 1}{x}\,dx$.

Solution.

(a) Write $3x^{-2} + x^{1/2} - 4 \cdot \tfrac{1}{x}$:

$$3 \cdot \frac{x^{-1}}{-1} + \frac{x^{3/2}}{3/2} - 4\ln|x| + C = -\frac{3}{x} + \frac{2}{3}x^{3/2} - 4\ln|x| + C$$

(b) Split the fraction: $\dfrac{x^2 + 1}{x} = x + \dfrac{1}{x}$.

$$\int \left(x + \frac{1}{x}\right)dx = \frac{x^2}{2} + \ln|x| + C$$

Example 3: Trig, exponential, and inverse trig

Find (a) $\displaystyle\int (2\cos x - \sec^2 x + 3e^x)\,dx$ and (b) $\displaystyle\int \frac{5}{1 + x^2}\,dx$.

Solution.

(a) $2\sin x - \tan x + 3e^x + C$

(b) $5\arctan x + C$

Check (a): $\dfrac{d}{dx}(2\sin x - \tan x + 3e^x) = 2\cos x - \sec^2 x + 3e^x$. ✓

Example 4: Initial-value problems

(a) Find $y$ if $\dfrac{dy}{dx} = 2x$ and $y(1) = 3$.

(b) Find $f(x)$ if $f''(x) = 6x - 2$, $f'(0) = 1$, and $f(0) = 4$.

Solution.

(a) $y = x^2 + C$. Substitute $x = 1$, $y = 3$: $3 = 1 + C$, so $C = 2$ and $y = x^2 + 2$. That’s the solid curve in the figure above.

(b) Integrate once:

$$f'(x) = 3x^2 - 2x + C_1, \qquad f'(0) = C_1 = 1$$

So $f'(x) = 3x^2 - 2x + 1$. Integrate again:

$$f(x) = x^3 - x^2 + x + C_2, \qquad f(0) = C_2 = 4$$ $$f(x) = x^3 - x^2 + x + 4$$

Common mistakes

Forgetting the + C. An indefinite integral is a whole family of functions. On a test (AP, IB, or your class), leaving off $+ C$ can cost a mark.

Using the power rule on 1/x. $\int x^{-1}\,dx$ is not $\tfrac{x^0}{0}$. It’s $\ln|x| + C$.

Getting trig signs backwards. The derivative of $\cos x$ is $-\sin x$, so $\int \sin x\,dx = -\cos x + C$, and $\int \cos x\,dx = +\sin x + C$. If in doubt, differentiate your answer.

Integrating products or quotients piece by piece. $\int x\sqrt{x}\,dx$ is not $\tfrac{x^2}{2} \cdot \tfrac{2}{3}x^{3/2}$. Rewrite as $\int x^{3/2}\,dx = \tfrac{2}{5}x^{5/2} + C$ first.

Solving for C too early or too late. Find $C$ right after integrating, before you integrate again. In Example 4(b), it’s easiest to find $C_1$ before the second integration.

Dividing by the old exponent. $\int x^3\,dx = \tfrac{x^4}{4} + C$, not $\tfrac{x^4}{3} + C$. Divide by the new exponent.

Practice

1. (Warm-up) Find $\displaystyle\int x^7\,dx$.

Solution

$$\frac{x^8}{8} + C$$

2. (Warm-up) Find $\displaystyle\int (4x^3 - 6)\,dx$.

Solution

$$x^4 - 6x + C$$

3. (Warm-up) Find $\displaystyle\int (\sin x + e^x)\,dx$.

Solution

$$-\cos x + e^x + C$$

4. (Core) Find $\displaystyle\int (2x + 1)^2\,dx$.

Solution

Expand first: $(2x + 1)^2 = 4x^2 + 4x + 1$.

$$\int (4x^2 + 4x + 1)\,dx = \frac{4x^3}{3} + 2x^2 + x + C$$

5. (Core) Find $\displaystyle\int \frac{x^3 - 2\sqrt{x}}{x}\,dx$.

Solution

Split the fraction: $\dfrac{x^3}{x} - \dfrac{2x^{1/2}}{x} = x^2 - 2x^{-1/2}$.

$$\int \left(x^2 - 2x^{-1/2}\right)dx = \frac{x^3}{3} - 2 \cdot \frac{x^{1/2}}{1/2} + C = \frac{x^3}{3} - 4\sqrt{x} + C$$

6. (Core) Find $\displaystyle\int \left(\frac{3}{x} + \frac{2}{1 + x^2} - \frac{1}{\sqrt{1 - x^2}}\right)dx$.

Solution

$$3\ln|x| + 2\arctan x - \arcsin x + C$$

7. (Core) Find $f(x)$ if $f'(x) = 3x^2 + 4\sin x$ and $f(0) = 2$.

Solution

$$f(x) = x^3 - 4\cos x + C$$

$f(0) = 0 - 4\cos 0 + C = -4 + C = 2$, so $C = 6$.

$$f(x) = x^3 - 4\cos x + 6$$

8. (Challenge) Find $f(x)$ if $f''(x) = 12x^2 + e^x$, $f'(0) = 3$, and $f(0) = 1$.

Solution

$f'(x) = 4x^3 + e^x + C_1$. Then $f'(0) = 1 + C_1 = 3$, so $C_1 = 2$.

$f(x) = x^4 + e^x + 2x + C_2$. Then $f(0) = 1 + C_2 = 1$, so $C_2 = 0$.

$$f(x) = x^4 + e^x + 2x$$

9. (Challenge) A particle moves along a line with acceleration $a(t) = 6t - 4$ m/s². Its initial velocity is $v(0) = -2$ m/s and its initial position is $x(0) = 5$ m. Find its position at $t = 2$ s.

Solution

Velocity is an antiderivative of acceleration:

$$v(t) = 3t^2 - 4t + C_1, \qquad v(0) = C_1 = -2$$

Position is an antiderivative of velocity:

$$x(t) = t^3 - 2t^2 - 2t + C_2, \qquad x(0) = C_2 = 5$$$$x(2) = 8 - 8 - 4 + 5 = 1$$

The particle is at $x = 1$ m when $t = 2$ s.