Question

using the general power rule find the indefinite integral.

∫[x^2]dx/(x^3-2)^2

Answer 1

∫[x^2]dx/(x^3-2)^2

Let x^3-2 =t

Take derivative

3x^2 =dt/dx

x^2dx =dt/3.

∫[x^2]dx/(x^3-2)^2 =∫dt/3t^2

                             =(1/3)∫dt/t^2

                             =(1/3)(-1/t)+C            (since ∫dt/t^2 =(t^(-2+1))/(-2+1))

                             =-1/(3(x^3-2))+C