# Integration of standard functions

## Summary and examples

In the following $F(x)$ is called the primitive of $f(x)$:

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

where $F(x)$ satisfies:

$\displaystyle\frac{dF}{dx}=f(x)$

 f(x) F(x) $x^n$ $\displaystyle\frac{1}{n+1}x^{n+1}$ a. $c$(constante) $cx$ b. $\sin(x)$ $-\cos(x)$ c. $\cos(x)$ $\sin(x)$ d. $e^x$ $e^x$ e. $\displaystyle\frac{1}{x}$ $\ln|x|$ f. $a^x$ $\displaystyle\frac{a^x}{\ln(a)}$ g.

Furthermore the following rules hold:

h. $\displaystyle\int{[af(x)+bg(x)]dx}=a\displaystyle\int{f(x)dx}+b\displaystyle\int{g(x)dx}$

Thus far the integrals are so-called indefinite integrals, meaning that the integral sign has no boundaries. This is in contrast to definite integrals which have an upper and a lower boundary.

For a definite integral we have:

i. $\displaystyle\int_{a}^{b}f(x)dx=[F(x)]_{a}^{b}=F(b)-F(a)$

In the following examples we show how we calculate integrals. The specific rule is also given.

##### Example 1

Solve:

$\displaystyle\int x^3dx$

The primitive function is calculated by applying a. and thus:

$\displaystyle\int x^3dx=\displaystyle\frac{1}{4}x^4+C$

##### Example 2

Solve:

$\displaystyle\int 5x^2dx$

The primitive is calculated by applying a. in combination with h. an thus:

$\displaystyle\int 5x^2dx=\displaystyle\frac{5}{3}x^3+C$

##### Example 3

Solve:

$\displaystyle\int 7dx$

The primitive is calculated by applying b. and thus:

$\displaystyle\int 7dx=7x+C$

##### Example 4

Solve:

$\displaystyle\int 8e^xdx$

The primitive is calculated by applying e. in combination met h. and thus:

$\displaystyle\int 8e^xdx=8e^x+C$

##### Example 5

Solve:

$\displaystyle\int_{1}^{3} \displaystyle\frac{9}{x}dx$

The primitive is calculated by applying f. in combination with h. and i. and thus:

$\displaystyle\int_{1}^{3} \displaystyle\frac{9}{x}dx=9[\ln|x|]_{1}^{3}=9[\ln(3)-\ln(1)]=9\ln(3)$

Remember, $\ln(1)=0$.

##### Example 6

Solve:

$\displaystyle\int_{0}^{1} (4x^2+6\cdot5^x)dx$

The primitive is calculated by applying a. and g. in combination with h. and i. and thus:

$\displaystyle\int_{0}^{1} (4x^2+6\cdot5^x)dx=4\displaystyle\int_{0}^{1}x^2dx+6\displaystyle\int_{0}^{1}5^xdx=$

$=[4\cdot\displaystyle\frac{1}{3}x^3+6\displaystyle\frac{5^x}{\ln(5)}]_{0}^{1}=\displaystyle\frac{4}{3}+\displaystyle\frac{6}{\ln(5)}=$

$\displaystyle\frac{4}{3}+\displaystyle\frac{24}{\ln(5)}$

##### Example 7

Solve:

$\displaystyle\int_{0}^{\pi} \sin(x)dx$

The primitive is calculated by applying c. in combination with i. and thus:

$\displaystyle\int_{0}^{\pi} \sin(x)dx=[-\cos(x)]_{0}^{\pi}=\cos(0)-\cos(\pi)=1-(-1)=2$

##### Example 8

Solve:

$5\displaystyle\int \sqrt[3]{x^2}dx$

This function does not seem to appear in the list of standard functions, but indirectly this is still the case. We can write:

$\sqrt[3]{x^2}=x^{\frac{2}{3}}$

and thus rule a. can be applied with $n=\displaystyle\frac{2}{3}$.

So:

$\displaystyle\int \sqrt[3]{x^2}dx=5\displaystyle\int {x^{\frac{2}{3}}}dx=\displaystyle\frac{5}{\frac{2}{3}+1}x^{\frac{2}{3}+1}+C=3x\sqrt[3]{x^2}+C$

##### Example 9

Solve:

$\displaystyle\int_{1}^{2} {\displaystyle\frac{9}{5x^3}}dx$

This function does not seem to appear in the list of standard functions, but indirectly this is still the case. We can write:

$\displaystyle\frac{9}{5x^3}=\displaystyle\frac{9}{5}x^{-3}$

and thus (by applying a., h. and i.):

$\displaystyle\int_{1}^{2} \displaystyle\frac{9}{5x^3}dx=$

$=\displaystyle\frac{9}{5}\displaystyle\int_{1}^{2} {x^{-3}}dx=$

$=[\displaystyle\frac{9}{5}\displaystyle\frac{1}{-3+1}x^{-3+1}]_{1}^{2}=$

$=[-\displaystyle\frac{9}{10x^2}]_{1}^{2}=$

=$\displaystyle\frac{27}{40}$

##### Example 10

Solve:

$\displaystyle\int \displaystyle\frac{5x^4+x}{x^4}dx$

This would seem an awkward integral, but the integrand (the function under the integral sign) can be written as follows:

$\displaystyle\frac{5x^4+x}{x^4}=\displaystyle\frac{5x^4}{x^4}+\displaystyle\frac{x}{x^4}=5+\displaystyle\frac{1}{x^3}=5+x^{-3}$

and thus the integral becomes:

$\displaystyle\int (5+x^{-3})dx=5x+\displaystyle\frac{1}{-2}x^{-2}+C=5x-\displaystyle\frac{1}{2x^2}+C$

Here a., b. and h. are applied.

##### Example 11

Solve:

$\displaystyle\int\displaystyle\frac{1}{\sqrt[5]{x^2}}dx$

We can write the integrand (the function under the integral sign):

$\displaystyle\frac{1}{\sqrt[5]{x^2}}=\displaystyle\frac{1}{x^{\frac{2}{5}}}=x^{-\frac{2}{5}}$

and thus the integral becomes (with a. and h.):

$\displaystyle\int{x^{-\frac{2}{5}}}dx=$

$=\displaystyle\frac{1}{-\frac{2}{5}+1}x^{-\frac{2}{5}+1}+C=$

$=\displaystyle\frac{5}{3}\sqrt[5]{x^3}+C$

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