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?

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 $
