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) dxxmeaxdx=xmeaxamaxm1eaxdx?

1 Answer
Dec 11, 2015

Set u = x^mu=xm and (dv)/dx = e^(ax)dvdx=eax and then apply the integration by parts formula.

Explanation:

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

intu(dv)/(dx)dx = uv-intv(du)/(dx)dxudvdxdx=uvvdudxdx

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


Starting from
intx^me^(ax)dxxmeaxdx

we set u(x) = x^mu(x)=xm and dv/dx = e^(ax)dvdx=eax.

Then, differentiating and integrating gives us
(du)/dx = mx^(m-1)dudx=mxm1 and v(x) = (e^(ax))/av(x)=eaxa

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)dxxmeaxdx=xmeaxaeaxamxm1dx

=(x^me^(ax))/a - m/a intx^(m-1)e^(ax)dx=xmeaxamaxm1eaxdx