$\displaystyle \int\frac{x^2}{1+({2x})^2} dx$ I know that the bottom is going to end up being $\displaystyle \arctan{2x}$ but how do I get rid of the $\displaystyle x^2$
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I would rewrite this and then do some algebra Originally Posted by summermagic $\displaystyle \int\frac{x^2}{1+4x^2} dx$ I know that the bottom is going to end up being $\displaystyle \arctan{2x}$ but how do I get rid of the $\displaystyle x^2$ Then $\displaystyle \int\frac{x^2}{1+4x^2} dx= (1/4)\int\frac{4x^2}{1+4x^2} dx= (1/4)\int\frac{(4x^2+1)-1}{1+4x^2} dx$ Then divide and integrate... $\displaystyle (1/4)\int\ \biggl(1-{1\over 1+4x^2}\biggr) dx$
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