How do you show that integration of $x^m e^(ax)dx = (x^m e^(ax) )/a - m/a int x^(m-1) e^(ax) dx$ ?

Question

4503 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-12-11T00:09:04

Set $u = x^m$ and $(dv)/dx = e^(ax)$ and then apply the integration by parts formula.

The integration by parts formula states that for continuous functions $u(x)$ and $v(x)$

$intu(dv)/(dx)dx = uv-intv(du)/(dx)dx$

(A short proof of the integration by parts formula can be found here)

Starting from
$intx^me^(ax)dx$

we set $u(x) = x^m$ and $dv/dx = e^(ax)$ .

Then, differentiating and integrating gives us
$(du)/dx = mx^(m-1)$ and $v(x) = (e^(ax))/a$

Substituting these into the integration by parts formula gives the desired result of
$intx^me^(ax)dx = x^m*e^(ax)/a - inte^(ax)/a*mx^(m-1)dx$

$=(x^me^(ax))/a - m/a intx^(m-1)e^(ax)dx$

How do you show that integration of x^m e^(ax)dx = (x^m e^(ax) )/a - m/a int x^(m-1) e^(ax) dxx^m e^(ax)dx = (x^m e^(ax) )/a - m/a int x^(m-1) e^(ax) dx?