# Which Basic Relations in Integration Should You Know?

When we say that integration is the same as “reverse derivation”, what we mean is that $\int f\left(x\right)\phantom{\rule{0.17em}{0ex}}dx$ is what you have to differentiate in order to get $f\left(x\right)$. So another word for integration is anti-derivation.

Even though integration might seem a bit cryptic in the beginning, the sums and differences of functions, and products and quotients of functions and numbers, all follow some simple rules. These rules show you how to integrate in these instances:

Rule

### UsefulIntegrationRules

$\begin{array}{llll}\hfill \int u\left(x\right)+v\left(x\right)\phantom{\rule{0.17em}{0ex}}dx& =\int u\left(x\right)\phantom{\rule{0.17em}{0ex}}dx+\int v\left(x\right)\phantom{\rule{0.17em}{0ex}}dx\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill \int u\left(x\right)-v\left(x\right)\phantom{\rule{0.17em}{0ex}}dx& =\int u\left(x\right)\phantom{\rule{0.17em}{0ex}}dx-\int v\left(x\right)\phantom{\rule{0.17em}{0ex}}dx\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

$\begin{array}{llll}\hfill \int ku\left(x\right)\phantom{\rule{0.17em}{0ex}}dx& =k\int u\left(x\right)\phantom{\rule{0.17em}{0ex}}dx\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill \int \frac{u\left(x\right)}{k}\phantom{\rule{0.17em}{0ex}}dx& =\frac{1}{k}\int u\left(x\right)\phantom{\rule{0.17em}{0ex}}dx\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

$\begin{array}{llll}\hfill \int u\left(x\right)+v\left(x\right)\phantom{\rule{0.17em}{0ex}}dx& =\int u\left(x\right)\phantom{\rule{0.17em}{0ex}}dx+\int v\left(x\right)\phantom{\rule{0.17em}{0ex}}dx\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill \int u\left(x\right)-v\left(x\right)\phantom{\rule{0.17em}{0ex}}dx& =\int u\left(x\right)\phantom{\rule{0.17em}{0ex}}dx-\int v\left(x\right)\phantom{\rule{0.17em}{0ex}}dx\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill \int ku\left(x\right)\phantom{\rule{0.17em}{0ex}}dx& =k\int u\left(x\right)\phantom{\rule{0.17em}{0ex}}dx\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill \int \frac{u\left(x\right)}{k}\phantom{\rule{0.17em}{0ex}}dx& =\frac{1}{k}\int u\left(x\right)\phantom{\rule{0.17em}{0ex}}dx\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

Example 1

Solve the integral $\int 2\mathrm{cos}\left(2x\right)+\frac{1}{x}\phantom{\rule{0.17em}{0ex}}dx$

$\begin{array}{llll}\hfill & \phantom{=}\int 2\mathrm{cos}\left(2x\right)+\frac{1}{x}\phantom{\rule{0.17em}{0ex}}dx\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =2\cdot \frac{1}{2}\mathrm{sin}\left(2x\right)+\mathrm{ln}\phantom{\rule{-0.17em}{0ex}}|x|+C\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\mathrm{sin}\left(2x\right)+\mathrm{ln}\phantom{\rule{-0.17em}{0ex}}|x|+C\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

$\begin{array}{llll}\hfill \int 2\mathrm{cos}\left(2x\right)+\frac{1}{x}\phantom{\rule{0.17em}{0ex}}dx& =2\cdot \frac{1}{2}\mathrm{sin}\left(2x\right)+\mathrm{ln}\phantom{\rule{-0.17em}{0ex}}|x|+C\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\mathrm{sin}\left(2x\right)+\mathrm{ln}\phantom{\rule{-0.17em}{0ex}}|x|+C\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

Example 2

Solve the integral $\int {e}^{3x}-\mathrm{sin}\left(\pi x\right)+\pi \phantom{\rule{0.17em}{0ex}}dx$

$\begin{array}{llll}\hfill & \phantom{=}\int {e}^{3x}-\mathrm{sin}\left(\pi x\right)+\pi \phantom{\rule{0.17em}{0ex}}dx\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\frac{1}{3}{e}^{3x}+\frac{1}{\pi }\mathrm{cos}\left(\pi x\right)+\pi x+C\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

 $\int {e}^{3x}-\mathrm{sin}\left(\pi x\right)+\pi \phantom{\rule{0.17em}{0ex}}dx=\frac{1}{3}{e}^{3x}+\frac{1}{\pi }\mathrm{cos}\left(\pi x\right)+\pi x+C$

Example 3

Solve the integral $\int 3{x}^{4}+3\mathrm{tan}\left(4x\right)\phantom{\rule{0.17em}{0ex}}dx$

$\begin{array}{llll}\hfill & \phantom{=}\int 3{x}^{4}+3\mathrm{tan}\left(4x\right)\phantom{\rule{0.17em}{0ex}}dx\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =3\frac{1}{5}{x}^{5}-3\cdot \frac{1}{4}\mathrm{ln}\phantom{\rule{-0.17em}{0ex}}|\mathrm{cos}\left(4x\right)|+C\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\frac{3}{5}{x}^{5}-\frac{3}{4}\mathrm{ln}\phantom{\rule{-0.17em}{0ex}}|\mathrm{cos}\left(4x\right)|+C\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

$\begin{array}{llll}\hfill \int 3{x}^{4}+3\mathrm{tan}\left(4x\right)\phantom{\rule{0.17em}{0ex}}dx& =3\frac{1}{5}{x}^{5}-3\cdot \frac{1}{4}\mathrm{ln}\phantom{\rule{-0.17em}{0ex}}|\mathrm{cos}\left(4x\right)|+C\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\frac{3}{5}{x}^{5}-\frac{3}{4}\mathrm{ln}\phantom{\rule{-0.17em}{0ex}}|\mathrm{cos}\left(4x\right)|+C\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

Example 4

Solve the integral $\int {2}^{x}-\frac{5}{3x+4}\phantom{\rule{0.17em}{0ex}}dx$

$\begin{array}{cc}\int {2}^{x}-\frac{5}{3x+4}\phantom{\rule{0.17em}{0ex}}dx& \\ =\frac{{2}^{x}}{\mathrm{ln}2}-5\cdot \frac{1}{3}\mathrm{ln}\phantom{\rule{-0.17em}{0ex}}|3x+4|+C& \\ =\frac{{2}^{x}}{\mathrm{ln}2}-\frac{5}{3}\mathrm{ln}\phantom{\rule{-0.17em}{0ex}}|3x+4|+C& \end{array}$

$\begin{array}{llll}\hfill \int {2}^{x}-\frac{5}{3x+4}\phantom{\rule{0.17em}{0ex}}dx& =\frac{{2}^{x}}{\mathrm{ln}2}-5\cdot \frac{1}{3}\mathrm{ln}\phantom{\rule{-0.17em}{0ex}}|3x+4|+C\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\frac{{2}^{x}}{\mathrm{ln}2}-\frac{5}{3}\mathrm{ln}\phantom{\rule{-0.17em}{0ex}}|3x+4|+C\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$