$\int \frac{x^3}{(x+1)^2} \mathrm{~d} x=$

To evaluate $\int \frac{x^3}{(x+1)^2} \mathrm{d}x$, use the substitution $u = x + 1$, which gives $x = u - 1$ and $dx = du$. The integral becomes $\int \frac{(u-1)^3}{u^2} du$.

Expanding the numerator: $(u-1)^3 = u^3 - 3u^2 + 3u - 1$, so the integrand simplifies to $\frac{u^3 - 3u^2 + 3u - 1}{u^2} = u - 3 + \frac{3}{u} - u^{-2}$.

Integrating term by term:

$\int \left(u - 3 + \frac{3}{u} - u^{-2}\right) du = \frac{u^2}{2} - 3u + 3\ln|u| + \frac{1}{u} + c$

Substituting back $u = x + 1$:

$I = \frac{(x+1)^2}{2} - 3(x+1) + 3\ln|x+1| + \frac{1}{x+1} + c$

Simplifying the algebraic terms:

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

Combining like terms: $\frac{x^2}{2} - 2x - \frac{5}{2} + 3\ln|x+1| + \frac{1}{x+1} + c$

The constant term $-\frac{5}{2}$ can be absorbed into the integration constant, yielding the final result:

$\frac{x^2}{2} - 2x + 3\ln|x+1| + \frac{1}{x+1} + C$

This matches option A.