Question

Evaluate the integral? : `int x^3e^(x^2) dx`

Answer 1

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

Explanation:

We need the integration by parts

`intp'qdx=pq-intpq'dx`

We perform the substitution

Let `u=x^2`, `=>`, `du=2xdx`

Therefore,

`intx^3e^(x^2)dx=1/2intue^ udu`

We apply the integration by parts

`p'(u)=e^u`, `=>`, `p(u)=e^u`

`q(u)=u`, `=>`, `q'(u)=1`

Therefore,

`1/2intue^ udu=1/2(ue^u-inte^udu)=1/2(ue^u-e ^u)`

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

Answer 2

We seek:

`int \ x^3e^(x^2) \ dx = 1/2(x^2-1)e^(x^2) + C`

Explanation:

We seek:

`I = int \ x^3e^(x^2) \ dx`

Note as a helper that:

`d/dx ( e^(x^2) ) = 2xe^(x^2) iff int \ 2xe^(x^2) \ dx = e^(x^2)`

So we can write the integral as:

`2I = int \ (x^2)(2xe^(x^2)) \ dx`

We can now use the formula for Integration By Parts (IBP):

`int \ u(dv)/dx \ dx = uv - int \ v(du)/dx \ dx`, or less formally
`" " int \ u \ dv=uv-int \ v \ du`

I was taught to remember the less formal rule in word; "The integral of udv equals uv minus the integral of vdu". If you struggle to remember the rule, then it may help to see that it comes a s a direct consequence of integrating the Product Rule for differentiation.

Let `{ (u,=x^2, => (du)/dx,=2x), ((dv)/dx,=2xe^(x^2), => v,=e^(x^2) ) :}`

Then plugging into the IBP formula:

`int \ (u)((dv)/dx) \ dx = (u)(v) - int \ (v)((du)/dx) \ dx`

gives us

`int \ (x^2)(2xe^(x^2)) \ dx = (x^2)(e^(x^2)) - int \ (e^(x^2))(2x) \ dx`

`:. 2I = x^2e^(x^2) - int \ 2xe^(x^2) \ dx + A`
`" " = x^2e^(x^2) - e^(x^2) + A`
`" " = (x^2-1)e^(x^2) + A`

Hence

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