# Last Integration Problem...I Swear!

• March 30th 2009, 10:49 PM
summermagic
Last Integration Problem...I Swear!
$\int\frac{x^2}{1+({2x})^2} dx$

I know that the bottom is going to end up being
$\arctan{2x}$

but how do I get rid of the $x^2$
• March 30th 2009, 10:54 PM
matheagle
I would rewrite this and then do some algebra

Quote:

Originally Posted by summermagic
$\int\frac{x^2}{1+4x^2} dx$

I know that the bottom is going to end up being
$\arctan{2x}$

but how do I get rid of the $x^2$

Then $\int\frac{x^2}{1+4x^2} dx= (1/4)\int\frac{4x^2}{1+4x^2} dx= (1/4)\int\frac{(4x^2+1)-1}{1+4x^2} dx$

Then divide and integrate...

$(1/4)\int\ \biggl(1-{1\over 1+4x^2}\biggr) dx$