How do you integrate #(5x-3)^2dx#?

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How do you integrate #(5x-3)^2dx#?

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Answer 1 · 2018-05-25T06:44:33

#int (5x-3)^2dx=1/15*(5x-3)^3+C#

Explanation:

Making the substitution
#t=5x-3# then #dx=1/5dt#
then we have
#1/5*int t^2dt#
using that #int x^ndx=x^(n+1)/(n+1)+C# with #n ne -1# then we obatin
#int (5x-3)^2dx=1/15*(5x-3)^3+C#

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How do you integrate #(5x-3)^2dx#?

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