Primitive of Power

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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.

$\blacksquare$


Sources