# Integrate x^2/(1+4•x^2)? Partial fractions

One last question

to Integrate x^2/(1+4•x^2). I would assume you would do long division but 4x^2 is bigger than x^2. so would you either pull out a 1/4 and it would be 1/4 ∫ x^2/(1/4+•x^2) dx or would the first term when doing long division be 1/4? or am I just totally wrong and you wouldn't do long division because the top is smaller than the bottom?

Yes, you do a long division and you get 1/4 + remainder. What is the remainder?

Yes, you do a long division and you get 1/4 + remainder. What is the remainder?

Ok, finished it but it was ugly. And I think I messed up my algebra somewhere.
your remainder ends up being -1/4 so you get∫ 1/4 +( -1/4/ (1+4x^2))so I decided to pull out the 4 to get the form x^2+a^2 so you get 1/4x + ∫-1/4 / (4((1/4)+x^2)) --> -1/16∫ 1/((1/4)+x^2) -->-1/16(1/(1/2)) times arctan(x/(1/2)) + C