Question

Question #a8dd2

Answer 1

`intx^3e^(2x^2)dx=1/8e^(2x^2)(2x^2-1)+C`

Explanation:

`I=intx^3e^(2x^2)dx`

Before using integration by parts (IBP), we can make a simpler substitution. Let `t=2x^2` so that `dt=4xdx`. This also implies that `x^2=1/2t`.

Modifying the integral:

`I=1/4intx^2e^(2x^2)(4xdx)=1/4int1/2te^tdt=1/8intte^tdt`

Now we should apply IBP. This integration technique takes the form `intudv=uv-intvdu`. So, for `intte^tdt`, we should let:

`{(u=t,=>,du=dt),(dv=e^tdt,=>,v=e^t):}`

Recall to differentiate `u` and integrate `dv`.

Thus:

`I=1/8(te^t-inte^tdt)=1/8(te^t-e^t)=1/8e^t(t-1)`

Since `t=2x^2` this becomes:

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