Antiderivative

Antidifferentiation (also called indefinite integration) is a thing done in mathematics. It is the opposite of differentiation.

Antiderivatives can tell you about size in a general way. Antidifferentiation is done on things like equations. Antidifferentiation gives you a thing called an antiderivative. An antiderivative is another kind of equation. Antidifferentiation is like integration with but without limits. This is why it is called indefinite.

An antiderivative is written like ${\displaystyle \int x\ dx}$

• The long S, ${\displaystyle \int }$, is called an integral sign. In integration, the integral sign has numbers on it. Those numbers tell you how to do the integration. Antiderivatives are different. They do not have numbers on on their integral signs.
• ${\displaystyle x}$is the equation you are integrating.
• The letters ${\displaystyle dx}$mean "with respect to ${\displaystyle x}$". This tells you how to do the antidifferentiation.

Simple integration

To do integrate ${\displaystyle ax^{n}}$

• Add 1 to the power ${\displaystyle n}$, so ${\displaystyle ax^{n}}$ is now ${\displaystyle ax^{n+1}}$
• Divide all this by the new power, so it is now ${\displaystyle {\frac {ax^{n+1}}{n+1}}}$
• Add constant ${\displaystyle c}$, so it is now ${\displaystyle {\frac {ax^{n+1}}{n+1}}+c}$

This can be shown as:

${\displaystyle \int ax^{n}\ dx={\frac {ax^{n+1}}{n+1}}+c}$

When there are many ${\displaystyle x}$ terms, integrate each part on its own:

${\displaystyle \int 2x^{6}-5x^{4}\ dx={\frac {2x^{7}}{7}}-{\frac {5x^{5}}{5}}+c={\frac {2}{7}}x^{7}-x^{5}+c}$

(This only works if the parts are being added or taken away.)

Examples

${\displaystyle \int 3x^{4}\ dx={\frac {3x^{5}}{5}}+c}$
${\displaystyle \int x+x^{2}+x^{3}+x^{4}\ dx={\frac {x^{2}}{2}}+{\frac {x^{3}}{3}}+{\frac {x^{4}}{4}}+{\frac {x^{5}}{5}}+c}$
${\displaystyle \int {\frac {1}{x+4}}\ dx=\ln |x+4|\times 1+c=\ln |x+4|+c}$

Changing fractions and roots into powers makes it easier:

${\displaystyle \int {\frac {1}{x^{3}}}\ dx=\int x^{-3}\ dx={\frac {x^{-2}}{-2}}+c=-{\frac {1}{2x^{2}}}+c}$
${\displaystyle \int {\sqrt {x^{3}}}\ dx=\int x^{\frac {3}{2}}\ dx={\frac {x^{\frac {5}{2}}}{\frac {5}{2}}}+c={\frac {2}{5}}x^{\frac {5}{2}}+c={\frac {2}{5}}{\sqrt {x^{5}}}+c}$

Integrating a bracket ("chain rule")

If you want to integrate a bracket like ${\displaystyle (2x+4)^{3}}$, we need to do it a different way. It is called the chain rule. It is like simple integration. It only works if the ${\displaystyle x}$ in the bracket has a power of 1 (it is linear) like ${\displaystyle x}$ or ${\displaystyle 5x}$ (not ${\displaystyle x^{5}}$ or ${\displaystyle x^{-7}}$).

To do ${\displaystyle \int (2x+4)^{3}\ dx}$

• Add 1 to the power ${\displaystyle 3}$, so that it is now ${\displaystyle (2x+4)^{4}}$
• Divide all this by the new power to get ${\displaystyle {\frac {(2x+4)^{4}}{4}}}$
• Divide all this by the derivative of the bracket ${\displaystyle \left({\frac {d(2x+4)}{dx}}=2\right)}$ to get ${\displaystyle {\frac {(2x+4)^{4}}{4\times 2}}={\frac {1}{8}}(2x+4)^{4}}$
• Add constant ${\displaystyle c}$ to give ${\displaystyle {\frac {1}{8}}(2x+4)^{4}+c}$

Examples

${\displaystyle \int (x+1)^{5}\ dx={\frac {(x+1)^{6}}{6\times 1}}+c={\frac {1}{6}}(x+1)^{6}+c\left(\because {\frac {d(x+1)}{dx}}=1\right)}$

${\displaystyle \int {\frac {1}{(7x+12)^{9}}}\ dx=\int (7x+12)^{-9}\ dx={\frac {(7x+12)^{-8}}{-8\times 7}}+c=-{\frac {1}{56}}(7x+12)^{-8}+c=-{\frac {1}{56(7x+12)^{8}}}+c\left(\because {\frac {d(7x+12)}{dx}}=7\right)}$