Integrate $\int \frac{(1-x)^2 \cdot e^x}{(1+x^2)^2}dx$ Integrate using integrating by parts : 

$$\int \frac{(1-x)^2 \cdot e^x}{(1+x^2)^2}dx$$

My attempt  :
$$I=\int \frac{(1-x)^2 \cdot e^x}{(1+x^2)^2}dx$$
$$I=\int \frac{e^x}{(1+x^2)^2}\cdot(1-x)^2dx$$
$$I=\frac{e^x}{(1+x^2)^2}\cdot\frac{(1-x)^3}{-3}-\int \frac{(1-x)^3}{-3}\cdot\frac{(1+x^2)^2e^x-e^x\cdot2(1+x^2)\cdot2x}{(1+x^2)^4}dx$$
Now it is little bit confusing. How can I simplify this ? Is there another method to do it differently ?

Hint
$$\frac{(1-x)^2 \cdot e^x}{(1+x^2)^2}=\frac{e^x}{1+x^2}-\frac{2xe^x}{(1+x^2)^2}$$
Indeed
$$(1-x)^2=(1+x^2)-2x$$
