Question

How do you integrate `x^3((x^4)+3)^5 dx`?

Answer 1

`(x^4+3)^6/24+C`

Explanation:

We have the integral

`intx^3(x^4+3)^5dx`

This can be solved fairly simply through substitution: we don't have to go through the hassle of expanding the binomial and then integrating term by term.

Let `u=x^4+3`. This implies that `du=4x^3dx`. Since we only have `x^3dx` in the integral and not `4x^3dx`, multiply the interior of the integral by `4` and the exterior by `1//4` to balance this.

`intx^3(x^4+3)^5dx=1/4int(x^4+3)^5*4x^3dx=1/4intu^5du`

Integrating this results in `(1/4)u^6/6+C`, and since `u=x^4+3`, this becomes `(x^4+3)^6/24+C`.