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

1 Answer

Nov 5, 2016

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

Explanation:

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#

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