How do you integrate $int x^2sqrt(x^3+1)$ using substitution?

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Answer 1 · 2017-06-25T19:39:42

The answer is $=2/9(x^3+1)^(3/2)+C$

We need

$intx^ndx=x^(n+1)/(n+1)+C(n!=-1)$

Perform the substitution

$u=x^3+1$

$du=3x^2dx$

$x^2dx=(du)/3$

Therefore,

$intx^2sqrt(x^3+1)dx=1/3intu^(1/2)du$

$=1/3u^(3/2)/(3/2)$

$=2/9u^(3/2)$

$=2/9(x^3+1)^(3/2)+C$

How do you integrate int x^2sqrt(x^3+1)int x^2sqrt(x^3+1) using substitution?