Reduction Formula for Power of x by General Exponential of a x

Theorem

$\ds \int x^n b^{a x} \rd x = \frac {x^n b^{a x} } {a \ln b} - \dfrac n {a \ln b} \int x^{n - 1} b^{a x} \rd x$

for $n \in \Z_{>0}$, $a \in \R_{\ne 0}$, $b > 0$, $b \ne 1$

With a view to expressing the primitive in the form:

$\ds \int u \frac {\d v} {\d x} \rd x = u v - \int v \frac {\d u} {\d x} \rd x$

let:

$\ds u$

$=$

$\ds x^n$

$\ds \leadsto \ \ $

$\ds \frac {\d u} {\d x}$

$=$

$\ds n x^{n - 1}$

and let:

$\ds \frac {\d v} {\d x}$

$=$

$\ds b^{a x}$

$\ds \leadsto \ \ $

$\ds v$

$=$

$\ds \dfrac {e^{a x} } {a \ln b}$

Then:

$\ds \int x^n b^{a x} \rd x$

$=$

$\ds \int u \rd v$

$\ds $

$=$

$\ds \paren {x^n} \paren {\dfrac {b^{a x} } {a \ln b} } - \int \paren {\dfrac {b^{a x} } {a \ln b} } \paren {n x^{n - 1} } \rd x$

$\ds $

$=$

$\ds \frac {x^n b^{a x} } {a \ln b} - \dfrac n {a \ln b} \int x^{n - 1} b^{a x} \rd x$

simplifying

$\blacksquare$