Question

How do you integrate `int x^2e^(4x)` by integration by parts method?

Answer 1

The answer is `=e^(4x)(x^2/4-x/8+1/32)+C`

Explanation:

Integration by parts is

`intuv'dx=uv-intu'vdx`

Here, we have

`u=x^2`, `=>`, `u'=2x`

`v'=e^(4x)`, `=>`, `v=e^(4x)/4`

Therefore,

`intx^2e^(4x)dx=(x^2e^(4x))/4-int2x*e^(4x)dx/4`

`=(x^2e^(4x))/4-1/2intx*e^(4x)dx`

We apply the integration by parts a second time to find `intx*e^(4x)dx`

`u=x`, `=>`, `u'=1`

`v'=e^(4x)`, `=>`, `v=e^(4x)/4`

Therefore,

`intx*e^(4x)dx=x/4e^(4x)-inte^(4x)dx/4`

`=x/4e^(4x)-e^(4x)/16`

Putting the results together,

`intx^2e^(4x)dx=(x^2e^(4x))/4-1/2(x/4e^(4x)-e^(4x)/16)`

`=e^(4x)(x^2/4-x/8+1/32)+C`

Answer 2

There is a faster method

Explanation:

Introduce `t` and write
`I = int e^(4tx) dx`
Now differentiate with respect to t
`(dI)/dt = int 4x e^(4tx) dx`
Once more
`(d^2I)/dt^2 = int 16x^2 e^(4tx) dx`
`I = e^(4tx)/(4t)`
`(dI)/dt= x/te^(4tx) - e^(4tx)/(4t^2)`
and again
`(d^2 I)/(dt^2) = ( 4x^2/t -x/(t^2)-x/t^2 + 2/t^3 1/4)e^(4tx)`
`(d^2I)/dt^2 = (4x^2-2x + 2)e^(4x)`
Our integral is 1/16 I
`J= I/16 = (x^2 - x/4 + 1/8)e^(4x)`