Computing Indefinite Integrals

Main idea

Every derivative rule you know, read backwards, is an integral rule. For powers that means raise the power by one and divide by the new power; the rest is a short table worth knowing by heart.

Power rule

\int x^{\htmlData{sym=0}{n}}\,dx = \frac{x^{{\htmlData{sym=0}{n}}+1}}{{\htmlData{sym=0}{n}}+1} + C \quad ({\htmlData{sym=0}{n}} \ne -1), \qquad \int\frac1x\,dx = \ln|x| + C

What the symbols mean

$n$
Exponent
Any real number except −1, which gives the log instead.

\int e^x dx = e^x + C, \quad \int\sin x\,dx = -\cos x + C, \quad \int\cos x\,dx = \sin x + C, \quad \int\sec^2 x\,dx = \tan x + C

\int\frac{dx}{1 + x^2} = \arctan x + C, \qquad \int\frac{dx}{\sqrt{1 - x^2}} = \arcsin x + C

Explanation

Computing indefinite integrals simply means finding antiderivatives of common functions quickly, using a few rules instead of guessing. All we're really doing is reading the derivative table from right to left. The power rule looks fussy, but it's two small moves: $x^{n+1}$ (raise the power by one) and $\frac{1}{n+1}$ (divide by the new power, to cancel the number that will come down when you differentiate). The exception, $n = -1$ , would mean dividing by zero, and that's where $\ln|x|$ steps in. The other formulas are derivatives you already know, flipped.

Think about why the power rule looks the way it does. Differentiating $x^4$ brings the 4 down and lowers the power: $4x^3$ . To undo that, you go the other way: raise the power back up to 4, then divide by 4 so the 4 that comes down cancels out. That's why $\int x^3\,dx = \frac{x^4}{4} + C$ .

Rewrite first, just like with derivatives: roots and fractions become powers.

\int\left(\sqrt x + \frac{1}{x^2}\right)dx = \int\left(x^{1/2} + x^{-2}\right)dx = \frac23x^{3/2} - x^{-1} + C.

A common mistake. It's tempting to write $\int\sin x\,dx = \cos x + C$ , because sine and cosine just seem to swap. It isn't: $(\cos x)' = -\sin x$ , so to get $+\sin x$ you need $-\cos x$ . That gives $\int\sin x\,dx = -\cos x + C$ . Differentiate your answer and the sign sorts itself out.

Initial value problems. If you also know one value of the function, like $f(1) = 5$ , plug it in after integrating to pin down $C$ .

See it

This is a car's speedometer reading, v(t) = 3 + 2 sin t, and its odometer, s(t) = 3t + 2 − 2 cos t, which is an antiderivative of v. As you can see, the slope of the odometer curve at each moment matches the speedometer reading, swinging between 1 and 5. Integrating the speed gives back the distance.

Worked examples

Example 1

Find $\displaystyle\int\left(5x^3 - 10x^{-6} + 4\right)dx$ .

1
Integrate term by term, raising each power by one and dividing by the new power: $\frac{5x^4}{4} - \frac{10x^{-5}}{-5} + 4x$ . The constant $4$ becomes $4x$ , since $(4x)' = 4$ .

$\frac54x^4 + 2x^{-5} + 4x + C$

Example 2

Find $f(x)$ if $f'(x) = 12x^2 - 4x$ and $f(-3) = 17$ .

1
Antidifferentiate: $f(x) = 4x^3 - 2x^2 + C$ .

Practice

Type your answer and check it. Solve them all to complete the lesson.

Problem 1+5 XP

Find $\displaystyle\int\left(6x^2 - 4x + 1\right)dx$ .

Problem 2+5 XP

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

Problem 3+5 XP

Find $\displaystyle\int\left(3e^x - 2\sin x\right)dx$ .

Problem 4+5 XP

Find $\displaystyle\int\frac{4}{1 + x^2}\,dx$ .

Problem 5+5 XP

Find $f(x)$ if $f'(x) = 6x - 4$ and $f(1) = 5$ .

Problem 6+5 XP

A city uses water at the rate $r(t) = 60 + 12t - t^2$ thousand liters per hour, $t$ hours after 6 AM. Let $W(t)$ be the total used since 6 AM, so $W' = r$ and $W(0) = 0$ . Find $W(6)$ , the total used by noon.

Challenge problem+25 XP

The hardest one here. Get it, and you've got this lesson.

Find $\displaystyle\int\frac{(x^2 + 1)^2}{\sqrt x}\,dx$ .

Where it's used

Given a rocket's acceleration and its launch speed, integrating once gives velocity; given the starting height too, integrating again gives altitude.

The Odometer and the SpeedometerCalc I · Physics

Indefinite Integrals The Substitution Rule