Question

How do you integrate by substitution `int x^3/sqrt(1+x^4) dx`?

Answer 1

`intx^3/sqrt(1+x^4)dx=sqrt(1+x^4)/2+C`

Explanation:

Using integration by substitution

let `u = 1+x^4 => du = 4x^3dx`

Then, together with the known formula `intx^ndx = x^(n+1)/(n+1)+C` for `n!=-1`, we have

`intx^3/sqrt(1+x^4)dx = 1/4int(4x^3)/sqrt(1+x^4)dx`

`=1/4int1/sqrt(u)du`

`=1/4intu^(-1/2)du`

`=1/4(u^(1/2)/(1/2))+C`

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

`=sqrt(1+x^4)/2+C`