What is the antiderivative of #(2x-1)^3#?

1 Answer
Adrian D.
Aug 6, 2015

#int(2x-1)^2dx=1/8(2x-1)^4+c#

Explanation:

To solve:
#int(2x-1)^2dx#

We use the substitution:
#u=2x-1# => #(du)/dx=2# => #dx=1/2du#

This gives:
#int(2x-1)^3dx=1/2intu^3du=1/8u^4+c#

Transforming back to the original variable we have:
#1/8u^4+c=1/8(2x-1)^4+c#

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