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#?

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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#