Indefinite Integrals

Notation

Definite and indefinite integrals share a symbol because of the Fundamental Theorem, but they are different objects. A definite integral

\int_a^b f(x)\,dx

is a number; an indefinite integral is a family of functions. For example,

\begin{gathered} \int 2x\,dx = x^2 + C, \\[6pt] \int_{1}^{3} 2x\,dx = 9 - 1 = 8. \end{gathered}

Basic integrals

Each rule below reverses a derivative formula and holds on any interval where the integrand is continuous. Where the domain has gaps, as for

\frac{1}{x}

, the constant can differ from one interval to the next.

Integral	Antiderivative
$\int k\,dx$	$kx + C$
$\int x^n\,dx$ , $n \ne -1$	$\dfrac{x^{n+1}}{n+1} + C$
$\int \dfrac{1}{x}\,dx$	$\ln\|x\| + C$
$\int e^x\,dx$	$e^x + C$
$\int a^x\,dx$ , $a > 0$ , $a \ne 1$	$\dfrac{a^x}{\ln a} + C$
$\int \cos x\,dx$	$\sin x + C$
$\int \sin x\,dx$	$-\cos x + C$
$\int \sec^2 x\,dx$	$\tan x + C$
$\int \csc^2 x\,dx$	$-\cot x + C$
$\int \sec x \tan x\,dx$	$\sec x + C$
$\int \csc x \cot x\,dx$	$-\csc x + C$
$\int \dfrac{1}{1 + x^2}\,dx$	$\arctan x + C$
$\int \dfrac{1}{\sqrt{1 - x^2}}\,dx$	$\arcsin x + C$

f

and

g

are continuous on an interval and

k

is a nonzero constant, then

\begin{aligned} &\int k f(x)\,dx = k\int f(x)\,dx, \\[6pt] &\int \bigl(f(x) \pm g(x)\bigr)\,dx \\[4pt] &\qquad = \int f(x)\,dx \pm \int g(x)\,dx. \end{aligned}

Example 1

Find each indefinite integral.

\begin{aligned} &\text{(a)}\;\; \int \left(3x^2 - \frac{4}{x} + 5\right)dx \\[10pt] &\text{(b)}\;\; \int \left(2\sec^2 x - \csc x \cot x\right)dx \\[10pt] &\text{(c)}\;\; \int \left(e^x + \frac{2}{1 + x^2}\right)dx \end{aligned}

(a) Term by term,

\begin{aligned} &\int \left(3x^2 - \frac{4}{x} + 5\right)dx \\[4pt] &\qquad = x^3 - 4\ln|x| + 5x + C. \end{aligned}

(b) An antiderivative of

\csc x \cot x

-\csc x

, so the minus sign in front of that term makes it

+\csc x

\begin{aligned} &\int \left(2\sec^2 x - \csc x \cot x\right)dx \\[4pt] &\qquad = 2\tan x + \csc x + C. \end{aligned}

(c) From the table,

\begin{aligned} &\int \left(e^x + \frac{2}{1 + x^2}\right)dx \\[4pt] &\qquad = e^x + 2\arctan x + C. \end{aligned}

Example 2

Evaluate each integral.

\begin{aligned} &\text{(a)}\;\; \int_{0}^{1} \frac{1}{1 + x^2}\,dx \\[10pt] &\text{(b)}\;\; \int_{0}^{1/2} \frac{1}{\sqrt{1 - x^2}}\,dx \end{aligned}

(a) The region under 1/(1 + x²) from 0 to 1 has area π/4.

(b) The region under 1/√(1 − x²) from 0 to 1/2 has area π/6.

(a) An antiderivative is

\arctan x

, so

\begin{aligned} &\int_{0}^{1} \frac{1}{1 + x^2}\,dx \\[4pt] &\qquad = \arctan 1 - \arctan 0 \\[4pt] &\qquad = \frac{\pi}{4}. \end{aligned}

(b) An antiderivative is

\arcsin x

, so

\begin{aligned} &\int_{0}^{1/2} \frac{1}{\sqrt{1 - x^2}}\,dx \\[4pt] &\qquad = \arcsin \frac{1}{2} - \arcsin 0 \\[4pt] &\qquad = \frac{\pi}{6}. \end{aligned}

Rewriting before integrating

The table has no general rule for a product, a quotient, or a power of a sum. The integral of a product isn't the product of the integrals. When the integrand multiplies out, divides term by term, or yields to an identity, rewrite it in a form the table covers.

Example 3

Find each indefinite integral.

\begin{aligned} &\text{(a)}\;\; \int \left(x + \frac{1}{x}\right)^2 dx \\[10pt] &\text{(b)}\;\; \int \frac{x^2 - 3x + 2}{x}\,dx \\[10pt] &\text{(c)}\;\; \int \sqrt{x}\,(x - 4)\,dx \end{aligned}

(a) Multiply out the square first:

\begin{aligned} &\int \left(x + \frac{1}{x}\right)^2 dx \\[4pt] &\qquad = \int \left(x^2 + 2 + x^{-2}\right)dx \\[4pt] &\qquad = \frac{x^3}{3} + 2x - \frac{1}{x} + C. \end{aligned}

(b) Divide each term of the numerator by

x

\begin{aligned} &\int \frac{x^2 - 3x + 2}{x}\,dx \\[4pt] &\qquad = \int \left(x - 3 + \frac{2}{x}\right)dx \\[4pt] &\qquad = \frac{x^2}{2} - 3x + 2\ln|x| + C. \end{aligned}

(c) Distribute

\sqrt{x} = x^{1/2}

, then use the power rule on each term:

\begin{aligned} &\int \sqrt{x}\,(x - 4)\,dx \\[4pt] &\qquad = \int \left(x^{3/2} - 4x^{1/2}\right)dx \\[4pt] &\qquad = \frac{2}{5}x^{5/2} - \frac{8}{3}x^{3/2} + C. \end{aligned}

Example 4

Find each indefinite integral.

\begin{aligned} &\text{(a)}\;\; \int \tan^2 x\,dx \\[10pt] &\text{(b)}\;\; \int \frac{\sin x}{\cos^2 x}\,dx \end{aligned}

(a) The table has no row for

\tan^2 x

, but the identity

\tan^2 x = \sec^2 x - 1

turns it into two rows that are there:

\begin{aligned} &\int \tan^2 x\,dx \\[4pt] &\qquad = \int \left(\sec^2 x - 1\right)dx \\[4pt] &\qquad = \tan x - x + C. \end{aligned}

(b) Split the fraction into two factors:

\frac{\sin x}{\cos^2 x} = \frac{1}{\cos x}\cdot\frac{\sin x}{\cos x} = \sec x \tan x

. So

\begin{aligned} &\int \frac{\sin x}{\cos^2 x}\,dx \\[4pt] &\qquad = \int \sec x \tan x\,dx \\[4pt] &\qquad = \sec x + C. \end{aligned}

A particular antiderivative

On an interval, the general antiderivative has one unknown constant. One given value

F(x_0) = y_0

, called an initial condition, determines that constant and picks out one member of the family, a particular antiderivative.