How do you integrate $int x^3/(1+x^4)^(1/3)$ using substitution?

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Answer 1 · 2016-10-05T02:34:42

$3/8(1+x^4)^(2/3) + C$

Let $1+x^4=t$ . Then $4x^3 dx = dt$ and the integral becomes

$1/4 int t^(-1/3) dt$

$=(1/4)t^(-1/3+1)/((-1/3+1)) + C$

$=3/8(1+x^4)^(2/3) + C$

How do you integrate int x^3/(1+x^4)^(1/3)int x^3/(1+x^4)^(1/3) using substitution?