Question

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`?

Answer 1

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

Explanation:

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)

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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`