# 4.10 Antiderivatives  (Page 3/10)

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Integration formulas
Differentiation Formula Indefinite Integral
$\frac{d}{dx}\left(k\right)=0$ $\int kdx=\int k{x}^{0}dx=kx+C$
$\frac{d}{dx}\left({x}^{n}\right)=n{x}^{n-1}$ $\int {x}^{n}dn=\frac{{x}^{n+1}}{n+1}+C$ for $n\ne \text{−}1$
$\frac{d}{dx}\left(\text{ln}|x|\right)=\frac{1}{x}$ $\int \frac{1}{x}dx=\text{ln}|x|+C$
$\frac{d}{dx}\left({e}^{x}\right)={e}^{x}$ $\int {e}^{x}dx={e}^{x}+C$
$\frac{d}{dx}\left(\text{sin}\phantom{\rule{0.1em}{0ex}}x\right)=\text{cos}\phantom{\rule{0.1em}{0ex}}x$ $\int \text{cos}\phantom{\rule{0.1em}{0ex}}x\phantom{\rule{0.1em}{0ex}}dx=\text{sin}\phantom{\rule{0.1em}{0ex}}x+C$
$\frac{d}{dx}\left(\text{cos}\phantom{\rule{0.1em}{0ex}}x\right)=\text{−}\text{sin}\phantom{\rule{0.1em}{0ex}}x$ $\int \text{sin}\phantom{\rule{0.1em}{0ex}}x\phantom{\rule{0.1em}{0ex}}dx=\text{−}\text{cos}\phantom{\rule{0.1em}{0ex}}x+C$
$\frac{d}{dx}\left(\text{tan}\phantom{\rule{0.1em}{0ex}}x\right)={\text{sec}}^{2}x$ $\int {\text{sec}}^{2}x\phantom{\rule{0.1em}{0ex}}dx=\text{tan}\phantom{\rule{0.1em}{0ex}}x+C$
$\frac{d}{dx}\left(\text{csc}\phantom{\rule{0.1em}{0ex}}x\right)=\text{−}\text{csc}\phantom{\rule{0.1em}{0ex}}x\phantom{\rule{0.1em}{0ex}}\text{cot}\phantom{\rule{0.1em}{0ex}}x$ $\int \text{csc}\phantom{\rule{0.1em}{0ex}}x\phantom{\rule{0.1em}{0ex}}\text{cot}\phantom{\rule{0.1em}{0ex}}x\phantom{\rule{0.1em}{0ex}}dx=\text{−}\text{csc}\phantom{\rule{0.1em}{0ex}}x+C$
$\frac{d}{dx}\left(\text{sec}\phantom{\rule{0.1em}{0ex}}x\right)=\text{sec}\phantom{\rule{0.1em}{0ex}}x\phantom{\rule{0.1em}{0ex}}\text{tan}\phantom{\rule{0.1em}{0ex}}x$ $\int \text{sec}\phantom{\rule{0.1em}{0ex}}x\phantom{\rule{0.1em}{0ex}}\text{tan}\phantom{\rule{0.1em}{0ex}}x\phantom{\rule{0.1em}{0ex}}dx=\text{sec}\phantom{\rule{0.1em}{0ex}}x+C$
$\frac{d}{dx}\left(\text{cot}\phantom{\rule{0.1em}{0ex}}x\right)=\text{−}{\text{csc}}^{2}x$ $\int {\text{csc}}^{2}x\phantom{\rule{0.1em}{0ex}}dx=\text{−}\text{cot}\phantom{\rule{0.1em}{0ex}}x+C$
$\frac{d}{dx}\left({\text{sin}}^{-1}x\right)=\frac{1}{\sqrt{1-{x}^{2}}}$ $\int \frac{1}{\sqrt{1-{x}^{2}}}={\text{sin}}^{-1}x+C$
$\frac{d}{dx}\left({\text{tan}}^{-1}x\right)=\frac{1}{1+{x}^{2}}$ $\int \frac{1}{1+{x}^{2}}dx={\text{tan}}^{-1}x+C$
$\frac{d}{dx}\left({\text{sec}}^{-1}|x|\right)=\frac{1}{x\sqrt{{x}^{2}-1}}$ $\int \frac{1}{x\sqrt{{x}^{2}-1}}dx={\text{sec}}^{-1}|x|+C$

From the definition of indefinite integral of $f,$ we know

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

if and only if $F$ is an antiderivative of $f.$ Therefore, when claiming that

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

it is important to check whether this statement is correct by verifying that ${F}^{\prime }\left(x\right)=f\left(x\right).$

## Verifying an indefinite integral

Each of the following statements is of the form $\int f\left(x\right)dx=F\left(x\right)+C.$ Verify that each statement is correct by showing that ${F}^{\prime }\left(x\right)=f\left(x\right).$

1. $\int \left(x+{e}^{x}\right)dx=\frac{{x}^{2}}{2}+{e}^{x}+C$
2. $\int x{e}^{x}dx=x{e}^{x}-{e}^{x}+C$
1. Since
$\frac{d}{dx}\left(\frac{{x}^{2}}{2}+{e}^{x}+C\right)=x+{e}^{x},$

the statement
$\int \left(x+{e}^{x}\right)dx=\frac{{x}^{2}}{2}+{e}^{x}+C$

is correct.
Note that we are verifying an indefinite integral for a sum. Furthermore, $\frac{{x}^{2}}{2}$ and ${e}^{x}$ are antiderivatives of $x$ and ${e}^{x},$ respectively, and the sum of the antiderivatives is an antiderivative of the sum. We discuss this fact again later in this section.
2. Using the product rule, we see that
$\frac{d}{dx}\left(x{e}^{x}-{e}^{x}+C\right)={e}^{x}+x{e}^{x}-{e}^{x}=x{e}^{x}.$

Therefore, the statement
$\int x{e}^{x}dx=x{e}^{x}-{e}^{x}+C$

is correct.
Note that we are verifying an indefinite integral for a product. The antiderivative $x{e}^{x}-{e}^{x}$ is not a product of the antiderivatives. Furthermore, the product of antiderivatives, ${x}^{2}{e}^{x}\text{/}2$ is not an antiderivative of $x{e}^{x}$ since
$\frac{d}{dx}\left(\frac{{x}^{2}{e}^{x}}{2}\right)=x{e}^{x}+\frac{{x}^{2}{e}^{x}}{2}\ne x{e}^{x}.$

In general, the product of antiderivatives is not an antiderivative of a product.

Verify that $\int x\phantom{\rule{0.1em}{0ex}}\text{cos}\phantom{\rule{0.1em}{0ex}}x\phantom{\rule{0.1em}{0ex}}dx=x\phantom{\rule{0.1em}{0ex}}\text{sin}\phantom{\rule{0.1em}{0ex}}x+\text{cos}\phantom{\rule{0.1em}{0ex}}x+C.$

$\frac{d}{dx}\left(x\phantom{\rule{0.1em}{0ex}}\text{sin}\phantom{\rule{0.1em}{0ex}}x+\text{cos}\phantom{\rule{0.1em}{0ex}}x+C\right)=\text{sin}\phantom{\rule{0.1em}{0ex}}x+x\phantom{\rule{0.1em}{0ex}}\text{cos}\phantom{\rule{0.1em}{0ex}}x-\text{sin}\phantom{\rule{0.1em}{0ex}}x=x\phantom{\rule{0.1em}{0ex}}\text{cos}\phantom{\rule{0.1em}{0ex}}x$

In [link] , we listed the indefinite integrals for many elementary functions. Let’s now turn our attention to evaluating indefinite integrals for more complicated functions. For example, consider finding an antiderivative of a sum $f+g.$ In [link] a. we showed that an antiderivative of the sum $x+{e}^{x}$ is given by the sum $\left(\frac{{x}^{2}}{2}\right)+{e}^{x}$ —that is, an antiderivative of a sum is given by a sum of antiderivatives. This result was not specific to this example. In general, if $F$ and $G$ are antiderivatives of any functions $f$ and $g,$ respectively, then

$\frac{d}{dx}\left(F\left(x\right)+G\left(x\right)\right)={F}^{\prime }\left(x\right)+{G}^{\prime }\left(x\right)=f\left(x\right)+g\left(x\right).$

Therefore, $F\left(x\right)+G\left(x\right)$ is an antiderivative of $f\left(x\right)+g\left(x\right)$ and we have

$\int \left(f\left(x\right)+g\left(x\right)\right)dx=F\left(x\right)+G\left(x\right)+C.$

Similarly,

$\int \left(f\left(x\right)-g\left(x\right)\right)dx=F\left(x\right)-G\left(x\right)+C.$

In addition, consider the task of finding an antiderivative of $kf\left(x\right),$ where $k$ is any real number. Since

$\frac{d}{dx}\left(kf\left(x\right)\right)=k\frac{d}{dx}F\left(x\right)=k{F}^{\prime }\left(x\right)$

for any real number $k,$ we conclude that

$\int kf\left(x\right)dx=kF\left(x\right)+C.$

These properties are summarized next.

## Properties of indefinite integrals

Let $F$ and $G$ be antiderivatives of $f$ and $g,$ respectively, and let $k$ be any real number.

Sums and Differences

$\int \left(f\left(x\right)\text{±}g\left(x\right)\right)dx=F\left(x\right)\text{±}G\left(x\right)+C$

Constant Multiples

$\int kf\left(x\right)dx=kF\left(x\right)+C$

From this theorem, we can evaluate any integral involving a sum, difference, or constant multiple of functions with antiderivatives that are known. Evaluating integrals involving products, quotients, or compositions is more complicated (see [link] b. for an example involving an antiderivative of a product.) We look at and address integrals involving these more complicated functions in Introduction to Integration . In the next example, we examine how to use this theorem to calculate the indefinite integrals of several functions.

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