The value of integral ∫e^x (1-x/1+x²)² dx=___?

In order to find the value of given integral \[\int e^{x}\left ( \frac{1-x}{1+x^{2}} \right )^{2} dx\].

For that consider property of integral: $\displaystyle \int e^{x}\left [ f(x)+{f}'(x) \right ] dx=e^{x} f(x)+C$ 

Hence

 \[\begin{align*} e^{x}\left ( \frac{1-x}{1+x^{2}} \right )^{2}&= e^{x}\frac{(1-x)^{2}}{(1+x^{2})^{2}}\\ &= e^{x}\left [ \frac{1-2x+x^{2}}{(1+x^{2})^{2}} \right ]\\ &= e^{x}\left [ \frac{1+x^{2}-2x}{(1+x^{2})^{2}} \right ]\\ &= e^{x}\left [ \frac{1+x^{2}}{(1+x^{2})^{2}}-\frac{2x}{(1+x^{2})^{2}} \right ]\\ &= e^{x}\left [ \frac{1}{(1+x^{2})}-\frac{2x}{(1+x^{2})^{2}} \right ]\\ \end{align*}\]

Here take $\displaystyle f(x)=\frac{1}{(1+x^{2})}\; \mathbf{and}\; {f}'(x)= \frac{-2x}{(1+x^{2})^{2}}$ and now by using property of integrals 

\[\begin{align*} \int e^{x}\left [ \frac{1-x }{1+x^{2}}\right ]^{2} dx&=\int e^{x}\left [ \frac{1}{(1+x^{2})}-\frac{2x}{(1+x^{2})^{2}} \right ]dx \\ &=\int e^{x}[f(x)+{f}'(x)]\; dx \\ &=e^{x} f(x) +C\\ &= e^{x} \frac{1}{1+x^{2}}+C \end{align*}\]

Hence $\displaystyle \begin{align*} \int e^{x}\left [ \frac{1-x }{1+x^{2}}\right ]^{2} dx &= \frac{e^{x}}{1+x^{2}}+C \end{align*}$ 

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