differential equation $y''-2y'+y=\frac{2(1-x^2) e^x}{(1+x^2)^2}$ using variation of parameter $y''-2y'+y=\frac{2(1-x^2) e^x}{(1+x^2)^2}$ 
so i got the homogeneous part
$y=Ae^x+Be^x . x$ 
i tried to find the non homogeneous part by using variation of parameter:
$W = \begin{vmatrix}
e^x& x.e^x  \\
e^x &e^x+x.e^x  \\
\end{vmatrix}$ =$e^{2x}$
$W_1=u_1'= \begin{vmatrix}
0& x.e^x  \\
\frac{2(1-x^2) e^x}{(1+x^2)^2}&e^x+x.e^x  \\
\end{vmatrix}$ =$\frac{-2x (1-x^2)}{(1+x^2)^2}$
$W_2=u_2' = \begin{vmatrix}
e^x& 0  \\
e^x&\frac{2(1-x^2) e^2x}{(1+x^2)^2} \\
\end{vmatrix}$ =$\frac{.2(1-x^2).}{(1+x^2)^2}$ 
$\int_{}^{} u_1'$$=\ln(x^2+1)+\frac{2}{x^2+1}$ 
however is my approach correct? for $\int u_2'$ the integral quite hard and i doubt i made mistake or maybe there is an easier method than this?  thanks so much!!

I think the idea is correct, but
$$u'_2=\frac{gy_1}W=\frac{2(1-x^2)e^x\cdot e^x}{e^{2x}(1+x^2)^2}=2\frac{1-x^2}{(1+x^2)^2}$$
and the integral of the above doesn't look so daunting.
Also
$$u'_1=-\frac{gy_2}W=-\frac{2(1-x^2)e^x\cdot xe^x}{e^{2x}(1+x^2)^2}=-2\frac{x(1-x^2)}{(1+x^2)^2}$$
and again: not so terrible an integral.
