$\int e^x \frac{(x-1)}{(x+1)^3} \mathrm{~d} x=$

Evaluate the integral $\int e^x \frac{x-1}{(x+1)^3} \mathrm{d} x$ using the standard formula $\int e^x [f(x) + f'(x)] \mathrm{d}x = e^x f(x) + C$.

Express the integrand in the appropriate form by rewriting the numerator:

$\frac{x-1}{(x+1)^3} = \frac{(x+1) - 2}{(x+1)^3} = \frac{1}{(x+1)^2} - \frac{2}{(x+1)^3}$

Let $f(x) = \frac{1}{(x+1)^2}$
Then $f'(x) = -\frac{2}{(x+1)^3}$
Thus the integrand becomes $f(x) + f'(x)$

Applying the formula yields:

$\int e^x \left[\frac{1}{(x+1)^2} - \frac{2}{(x+1)^3}\right] \mathrm{d} x = e^x \cdot \frac{1}{(x+1)^2} + C = \frac{e^x}{(x+1)^2} + C$

This corresponds to option C.