# Need help integrating $\frac{(t-1)^2-2t(t-1)}{t^2+(a(t-1)^2)^2}$

I need to integrate $$a\int_1^2 \frac{(t-1)^2-2t(t-1)}{t^2+(a(t-1)^2)^2} dt$$ $$=a\int_1^2 \frac{-t^2+1}{t^2+(a(t-1)^2)^2} dt$$ $$=-a\int_1^2 \frac{t^2-1}{t^2+(a(t-1)^2)^2} dt$$ with the hint that two trigonomic substitutions would be necessary and to consider using arctan. I have tried to tackle this a few different ways and get stuck every time. Can anyone help me start off?

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Maybe you can simplify this a bit first? –  ᴊ ᴀ s ᴏ ɴ Feb 27 '13 at 15:56
I think the key to the bottom is the two squares, so I'm wary of simplifying. –  Carly Feb 27 '13 at 15:58
Seems interessting, Mathematica needs more than minute (whithout a result) –  Dominic Michaelis Feb 27 '13 at 16:22

Looks like we could make use of this $$a\int \frac{-t^2+1}{t^2+(a(t-1)^2)^2} \; dt = -a \int \frac{1 - \frac{1}{t^2}}{1 + \left( a \left( \sqrt{t } - \frac{1}{\sqrt t } \right )^2 \right )^2 }dt = -a \int \frac{1 - \frac{1}{t^2}}{1 + \left( a \left( t + \frac{1}{ t } - 2 \right ) \right )^2 }dt$$
Substitute $\displaystyle a\left( t + \frac{1}{ t } - 2 \right ) = u$, you get $\displaystyle - \int \frac{ 1}{1 + (u)^2} du$