# What Are the Important Integration Rules?

Integration is to derivation as a plus is to a minus. That means that just like a minus reverses what a plus does, and vice versa, integration reverses what derivation does, and vice versa. Another way of putting it is that integration is finding the anti-derivative.

Now that you know this, you have also learned a method to check that your integrals are correct: After integrating, you can differentiate your answer to see whether you get the original expression that you integrated (called the integrand, the expression between the integral sign and $\phantom{\rule{0.17em}{0ex}}dx$).

Below is an overview of the integrals of several known and important functions. You will be expected to know these integral rules by heart. I recommend that you memorize them to learn them well, and that you always write down the proper formula before applying it when solving exercises.

Formula

### IntegrationFormulas

$\begin{array}{llll}\hfill & \int k\phantom{\rule{0.17em}{0ex}}dx=kx+C\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & \int {x}^{n}\phantom{\rule{0.17em}{0ex}}dx=\frac{1}{n+1}{x}^{n+1}+C,\phantom{\rule{0.33em}{0ex}}n\ne -1\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & \int \frac{1}{x}\phantom{\rule{0.17em}{0ex}}dx=\mathrm{ln}\phantom{\rule{-0.17em}{0ex}}|x|+C,\phantom{\rule{0.33em}{0ex}}x\ne 0\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & \int \mathrm{ln}kx\phantom{\rule{0.17em}{0ex}}dx=x\mathrm{ln}kx-x+C\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & \int {e}^{x}\phantom{\rule{0.17em}{0ex}}dx={e}^{x}+C\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & \int {e}^{kx}\phantom{\rule{0.17em}{0ex}}dx=\frac{1}{k}{e}^{kx}+C\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & \int \mathrm{cos}x\phantom{\rule{0.17em}{0ex}}dx=\mathrm{sin}x+C\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & \int \mathrm{cos}kx\phantom{\rule{0.17em}{0ex}}dx=\frac{1}{k}\mathrm{sin}kx+C\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & \int \mathrm{sin}x\phantom{\rule{0.17em}{0ex}}dx=-\mathrm{cos}x+C\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & \int \mathrm{sin}kx\phantom{\rule{0.17em}{0ex}}dx=-\frac{1}{k}\mathrm{cos}kx+C\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & \int \mathrm{tan}x\phantom{\rule{0.17em}{0ex}}dx=-\mathrm{ln}\phantom{\rule{-0.17em}{0ex}}|\mathrm{cos}x|+C,\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & \phantom{\int \mathrm{tan}x\phantom{\rule{0.17em}{0ex}}dx=}x\ne \frac{\pi }{2}+n\pi \phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & \int \mathrm{tan}kx\phantom{\rule{0.17em}{0ex}}dx=-\frac{1}{k}\mathrm{ln}\phantom{\rule{-0.17em}{0ex}}|\mathrm{cos}kx|+C\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & \int {a}^{x}\phantom{\rule{0.17em}{0ex}}dx=\frac{{a}^{x}}{\mathrm{ln}a}+C\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & \int \frac{1}{kx+a}\phantom{\rule{0.17em}{0ex}}dx=\frac{1}{k}\mathrm{ln}\phantom{\rule{-0.17em}{0ex}}|kx+a|+C\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

$\begin{array}{llll}\hfill & \int k\phantom{\rule{0.17em}{0ex}}dx=kx+C\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & \int {x}^{n}\phantom{\rule{0.17em}{0ex}}dx=\frac{1}{n+1}{x}^{n+1}+C,\phantom{\rule{0.33em}{0ex}}n\ne -1\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & \int \frac{1}{x}\phantom{\rule{0.17em}{0ex}}dx=\mathrm{ln}\phantom{\rule{-0.17em}{0ex}}|x|+C,\phantom{\rule{0.33em}{0ex}}x\ne 0\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & \int \mathrm{ln}kx\phantom{\rule{0.17em}{0ex}}dx=x\mathrm{ln}kx-x+C\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & \int {e}^{x}\phantom{\rule{0.17em}{0ex}}dx={e}^{x}+C\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & \int {e}^{kx}\phantom{\rule{0.17em}{0ex}}dx=\frac{1}{k}{e}^{kx}+C\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & \int \mathrm{cos}x\phantom{\rule{0.17em}{0ex}}dx=\mathrm{sin}x+C\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & \int \mathrm{cos}kx\phantom{\rule{0.17em}{0ex}}dx=\frac{1}{k}\mathrm{sin}kx+C\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & \int \mathrm{sin}x\phantom{\rule{0.17em}{0ex}}dx=-\mathrm{cos}x+C\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & \int \mathrm{sin}kx\phantom{\rule{0.17em}{0ex}}dx=-\frac{1}{k}\mathrm{cos}kx+C\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & \int \mathrm{tan}x\phantom{\rule{0.17em}{0ex}}dx=-\mathrm{ln}\phantom{\rule{-0.17em}{0ex}}|\mathrm{cos}x|+C,\phantom{\rule{0.33em}{0ex}}x\ne \frac{\pi }{2}+n\cdot \pi \phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & \int \mathrm{tan}kx\phantom{\rule{0.17em}{0ex}}dx=-\frac{1}{k}\mathrm{ln}\phantom{\rule{-0.17em}{0ex}}|\mathrm{cos}kx|+C\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & \int {a}^{x}\phantom{\rule{0.17em}{0ex}}dx=\frac{{a}^{x}}{\mathrm{ln}a}+C\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & \int \frac{1}{kx+a}\phantom{\rule{0.17em}{0ex}}dx=\frac{1}{k}\mathrm{ln}\phantom{\rule{-0.17em}{0ex}}|kx+a|+C\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

Note! The constant $C$ is not as mysterious as it might seem. It represents a possible constant that disappears when you differentiate. Because we don’t know what the constant is, $C$ can represent any number.