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!