Question

How do you integrate `int (3x^2)/sqrt(x^3+7)` using substitution?

Answer 1

`2sqrt(x^3+7)+C`

Explanation:

`int(3x^2)/sqrt(x^3+7)dx`

Apply the substitution `x^3+7=u`. Thus, `3x^2dx=du`. Substituting in:

`=int(du)/sqrtu`

`=intu^(-1/2)du`

Using the rule `intu^ndu=u^(n+1)/(n+1)+C`, where `n!=-1`:

`=u^(1/2)/(1/2)+C`

`=2sqrtu+C`

`=2sqrt(x^3+7)+C`