$\int e^x \left(\frac{1-x}{1+x^2}\right)^2 dx$ is equal to

The integral to solve is $\int e^x \left(\frac{1-x}{1+x^2}\right)^2 dx$

For integrals with $e^x$, the formula $\int e^x [f(x) + f'(x)] dx = e^x \cdot f(x) + C$ can be applied.

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Expanding the square:

$\left(\frac{1-x}{1+x^2}\right)^2 = \frac{(1-x)^2}{(1+x^2)^2}$

The integral becomes:

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

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Let $f(x) = \frac{1}{1+x^2}$

Finding the derivative:

$f'(x) = \frac{-2x}{(1+x^2)^2}$

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Adding $f(x)$ and $f'(x)$:

$f(x) + f'(x) = \frac{1}{1+x^2} + \frac{-2x}{(1+x^2)^2}$

$= \frac{(1+x^2)}{(1+x^2)^2} + \frac{-2x}{(1+x^2)^2}$

$= \frac{1+x^2-2x}{(1+x^2)^2}$

$= \frac{1-2x+x^2}{(1+x^2)^2}$

$= \frac{(1-x)^2}{(1+x^2)^2}$

This matches the integrand.

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Since $\frac{(1-x)^2}{(1+x^2)^2} = f(x) + f'(x)$ where $f(x) = \frac{1}{1+x^2}$:

$\int e^x \left[\frac{(1-x)^2}{(1+x^2)^2}\right] dx = e^x \cdot f(x) + C$

$= \frac{e^x}{1+x^2} + C$