How do you integrate by substitution $int x^2(x^3-1)^4 dx$ ?

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Answer 1 · 2016-11-05T07:36:50

THe answer is $(x^3-1)^5/15+C$

Let $u=(x^3-1)$
Then $du=3x^2dx$ $=>$ $(du)/3=x^2dx$
So $intx^2(x^3-1)^4dx=int(u^4du)/3$
$=1/3*u^5/5=u^5/15$
$:.intx^2(x^3-1)^4dx=(x^3-1)^5/15+C$

How do you integrate by substitution int x^2(x^3-1)^4 dxint x^2(x^3-1)^4 dx?