integral with parameter and 2 denominators Hello  I want to solve following integral 
$\displaystyle\int \frac{x^22t}{(1+x^2)(1+t^2x^2)}\,dx=-\dfrac{2\arctan(tx)-t \arctan(x)}{t^2-1}$
My Problem is that I just do not know how to start solving this problem .
 A: Partial fraction decomposition gives$$\frac 1{(1-t^2)(1+t^2x^2)}-\frac 1{(1-t^2)(1+x^2)}=\frac {x^2}{(1+x^2)(1+t^2x^2)}$$Hence, the integral now becomes$$\begin{align*}\mathfrak{I} & =2t\int\mathrm dx\,\left[\frac 1{(1-t^2)(1+t^2x^2)}-\frac 1{(1-t^2)(1+x^2)}\right]\\ & =\frac {2t}{1-t^2}\int\frac {\mathrm dx}{1+t^2x^2}-\frac {2t}{1-t^2}\int\frac {\mathrm dx}{1+x^2}\end{align*}$$Can you continue?
EDIT: The partial fraction decomposition can be derived by assuming the fraction splits into two separate components $$\frac {x^2}{(1+x^2)(1+t^2x^2)}=\frac {Ax+B}{1+x^2}+\frac {Cx+D}{1+t^2x^2}$$Now multiply both sides by $(1+x^2)(1+t^2x^2)$ to clear the fractions completely. Hence$$x^2=(Ax+B)(1+t^2x^2)+(Cx+D)(1+x^2)$$
To find the coefficients, we evaluate the equation first at $x^2=-1$ and then $x^2=-\tfrac 1{t^2}$. The aim of these substitutions is to set one of the products equal to zero and compare the coefficients to each other. By setting $x^2=-1$, then$$\begin{align*}-1 & =(Ax+B)(1-t^2)\\ & =Ax(1-t^2)+B(1-t^2)\end{align*}$$From here, it's evident that $A=0$ because there's no linear term in the left - hand side. Similarly, $B=-\tfrac 1{1-t^2}$. Repeating the process similarly for $x^2=-\tfrac 1{t^2}$ to get that $C=0$ and $D=\tfrac 1{1-t^2}$.$$\frac {x^2}{(1+x^2)(1+t^2x^2)}\color{red}{=\frac 1{(1-t^2)(1+t^2x^2)}-\frac 1{(1-t^2)(1+x^2)}}$$which aligns with what we have above.
A: Hint: Write your integrand in the form
$$\frac{1}{-(t-1)t^2(t+1)x^2-(t-1)(t+1)}+\frac{1}{(t-1)(tg+1)x^2+(t-1)(t+1)}$$
