Which solution to $\int \frac{x^3}{(x^2+1)^2}dx$ is correct? I tried solving the following integral using integral by parts :
$$\int \frac{x^3}{(x^2+1)^2}dx$$
but I got a different answer from Wolfram Calculator This is the answer that I got :
$$\int \frac{x^3}{(x^2+1)^2}dx=\frac{1}{2(x^2+1)}+\frac{1}{2}\ln(x^2+1)+C$$
I am wondering if I am wrong or the calculator if I am where did I do something wrong.

Here is my solution :
$$\int \frac{x^3}{(x^2+1)^2}dx$$
$$\int x^2\frac{x}{(x^2+1)^2}dx$$
$\implies u =x^2 \qquad u'=2x$
$\displaystyle\implies v'=\frac{x}{(x^2+1)^2}\qquad v=\int\frac{x}{(x^2+1)^2}dx$
$$\int\frac{x}{(x^2+1)^2}dx$$
$\implies \zeta=x^2+1$
$\implies\displaystyle \frac{d\zeta}{2}=x$
$$\boxed{v=\frac{1}{2}\int\frac{d\zeta}{\zeta^2}=-0.5\zeta^{-1}=\frac{-1}{2(x^2+1)}}$$
$$\int x^2\frac{x}{(x^2+1)^2}dx=x^2\frac{-1}{2(x^2+1)}-\underbrace{(\int \frac{-2x}{2(x^2+1)}dx)}_{-\frac{1}{2}\ln(x^2+1)}$$
$$\boxed{\int \frac{x^3}{(x^2+1)^2}dx=\frac{-x^2}{2(x^2+1)}+\frac{1}{2}\ln(x^2+1)+C}$$

BOTH are correct since the two answers differ by a constant.
$$
\begin{aligned}
\int \frac{x^{3}}{\left(x^{2}+1\right)^{2}} d x &=-\frac{1}{2} \int x^{2} d\left(\frac{1}{x^{2}+1}\right) \\
&=-\frac{x^{2}}{2\left(x^{2}+1\right)}+\frac{1}{2} \int \frac{2 x}{x^{2}+1} d x \\
&=-\frac{x^{2}}{2\left(x^{2}+1\right)}+\frac{1}{2} \ln \left(x^{2}+1\right)+C_{1} \\
&=-\frac{x^{2}+1-1}{2\left(x^{2}+1\right)}+\frac{1}{2} \ln \left(x^{2}+1\right)+C_{1} \\
&=-\frac{1}{2}+\frac{1}{2\left(x^{2}+1\right)}+\frac{1}{2} \ln \left(x^{2}+1\right)+C_{1}  \\
&=\frac{1}{2\left(x^{2}+1\right)}+\frac{1}{2} \ln \left(x^{2}+1\right)+C_{2},
\end{aligned}
$$
where $C_{2}= C_{1} -\frac{1}{2}$.
