Antiderivative

Antidifferentiation (or indefinite integration) is a part of mathematics. It is the opposite of differentiation. It is integrating with no limits. The answer is an equation.

It is written as $\int x\ dx$

• with the integral sign that has no limits $\int$
• the equation you are integrating $x$
• and the $dx$ which means "with respect to $x$", which does not mean anything with simple integration.

Simple integration

To integrate $ax^{n}$

• add 1 to the power $n$, so $ax^{n}$ is now $ax^{n+1}$
• divide all this by the new power, so it is now ${\frac {ax^{n+1}}{n+1}}$
• and a constant $c$ should be added, so it is now ${\frac {ax^{n+1}}{n+1}}+c$

This can be shown as:

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

When there are many $x$ terms, integrate each part on its own:

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

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

Examples

$\int 3x^{4}\ dx={\frac {3x^{5}}{5}}+c$

$\int x+x^{2}+x^{3}+x^{4}\ dx={\frac {x^{2}}{2}}+{\frac {x^{3}}{3}}+{\frac {x^{4}}{4}}+{\frac {x^{5}}{5}}+c$

$\int {\frac {1}{x+4}}\ dx=\ln(x+4)\times 1+c=\ln(x+4)+c$

Changing fractions and roots into powers makes it easier:

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

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

Integrating a bracket ("chain rule")

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

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

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

Examples

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

$\int {\frac {1}{(7x+12)^{9}}}\ dx=\int (7x+12)^{-9}\ dx={\frac {(7x+12)^{-8}}{-8\times 7}}+c=-{\frac {1}{54}}(7x+12)^{-8}+c=-{\frac {1}{54(7x+12)^{8}}}+c\left(\because {\frac {d(7x+12)}{dx}}=7\right)$

See also

• Mathematics
• Integral