<s>[INST] 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 ?
 [/INST] Just if you look for another way to do it.
Because of the square in denominator and the exponential in numerator, you could have assumed that the result is 
$$\int \frac{(1-x)^2 }{(1+x^2)^2}e^x\,dx= \frac {P_n(x)}{1+x^2}e^x$$ where $P_n(x)$ is a polynomial of degree $n$.
Differentiate both sides 
$$\frac{(1-x)^2 }{(1+x^2)^2}e^x=\frac{ \left(x^2+1\right) P'_n(x)+(1-x)^2 P_n(x)}{\left(1+x^2\right)^2}e^x$$ that is to say $$(1-x)^2=\left(x^2+1\right) P'_n(x)+(1-x)^2 P_n(x)$$ Comparing the degrees : $2$ for the lhs and $n+2$ in the rhs makes $n=0$; so $P_n(x)$ is just a constant. This implies that $P'_n(x)=0$ and what is left is $$(1-x)^2=(1-x)^2 P_0(x)$$ which makes $P_0(x)=1$.
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