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

Question

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

1 Answer

Impact of this question

5678 views around the world

You can reuse this answer
Creative Commons License

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#

Science

Math

Humanities

... and beyond

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

1 Answer

Explanation:

Related questions

Impact of this question