How do you find the antiderivative of $int x^2/sqrt(1-x^3)dx$ ?

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Answer 1 · 2017-01-14T01:09:22

$-2/3sqrt(1-x^3)+C$

Rewriting the function:

$I=intx^2/sqrt(1-x^3)dx=intx^2(1-x^3)^(-1/2)dx$

To deal with the $-1//2$ power, let $u=1-x^3$ . This implies that $du=-3x^2color(white).dx$ . Rewriting the integral:

$I=-1/3int(1-x^3)^(-1/2)(-3x^2color(white).dx)=-1/3intu^(-1/2)color(white).du$

Now use the rule $intu^ncolor(white).du=u^(n+1)/(n+1)$ :

$I=-1/3(u^(1/2)/(1/2))=-2/3sqrtu=-2/3sqrt(1-x^3)+C$

How do you find the antiderivative of int x^2/sqrt(1-x^3)dxint x^2/sqrt(1-x^3)dx?