# Primitive of Power

## Theorem

Let $n \in \R: n \ne -1$.

Then:

$\displaystyle \int x^n \rd x = \frac {x^{n + 1} } {n + 1} + C$

where $C$ is an arbitrary constant.

That is:

$\dfrac {x^{n + 1} } {n + 1}$ is a primitive of $x^n$.

## Proof

 $\displaystyle \frac \d {\d x} \paren {\dfrac {x^{n + 1} } {n + 1} }$ $=$ $\displaystyle \paren {n + 1} \paren {\dfrac {x^{\paren {n + 1} - 1} } {n + 1} }$ Power Rule for Derivatives $\displaystyle$ $=$ $\displaystyle x^n$ $\displaystyle \leadsto \ \$ $\displaystyle \int x^n \rd x$ $=$ $\displaystyle \frac {x^{n + 1} } {n + 1} + C$ Definition of Primitive (Calculus)

When $n = -1$ we have $n + 1 = 0$, and $\dfrac {x^{n + 1} } {n + 1} = \dfrac {x^0} 0$ is undefined.

$\blacksquare$