How do you integrate #((x^2) / (16-x^3)^2) dx#?

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How do you integrate #((x^2) / (16-x^3)^2) dx#?

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Answer 1 · 2015-09-13T21:13:35

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

Explanation:

Choose #t=16-x^3#, then
#dt=-3x^2dx => x^2dx=-dt/3#
#intx^2/(16-x^3)^2dx=int1/t^2(-dt/3)=-1/3intt^(-2)dt=#
#=-1/3 t^(-1)/-1+C=1/3 1/t+C=1/(3(16-x^3))+C#

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How do you integrate #((x^2) / (16-x^3)^2) dx#?

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