How do you integrate by substitution $int x^2/(16-x^3)^2 dx$ ?

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Answer 1 · 2017-02-08T14:35:44

$1/(3(16 - x^3)) + C$

Let $u = 16 - x^3$ . Then $du = -3x^2dx$ and $dx= (du)/(-3x^2)$ .

$=>int x^2/u^2 * (du)/(-3x^2)$

$=>-1/3int1/u^2 du$

$=>-1/3int u^-2du$

$=>-1/3(-u^-1) + C$

$=>1/(3u) + C$

$=>1/(3(16 - x^3)) + C$

Hopefully this helps!

How do you integrate by substitution int x^2/(16-x^3)^2 dxint x^2/(16-x^3)^2 dx?