# Primitive of Power of a x + b/Proof 2

## Theorem

$\ds \int \paren {a x + b}^n \rd x = \frac {\paren {a x + b}^{n + 1} } {\paren {n + 1} a} + C$

where $n \ne 1$.

## Proof

Let $u = a x + b$.

Then:

$\dfrac {\d u} {\d x} = a$

Then:

 $\ds \int \paren {a x + b}^n \rd x$ $=$ $\ds \int \dfrac {u^n} a \rd u$ Integration by Substitution $\ds$ $=$ $\ds \dfrac 1 a \dfrac {u^{n + 1} } {n + 1}$ Primitive of Power $\ds$ $=$ $\ds \frac {\paren {a x + b}^{n + 1} } {\paren {n + 1} a} + C$ substituting back for $u$

$\blacksquare$