<s>[INST] Calculating $\int \frac{dx}{(x^2+1)^3}$ integral Would anyone help me calculate the following integral? $\int \frac{dx}{(x^2+1)^3}$
During our lecutre we've done very similiar one, $\int \frac{dx}{(x^2+1)^2}$ like that: 
$\int \frac{dx}{(x^2+1)^2} = \int \frac{x^2+1-x^2}{(x^2+1)^2}dx = \int \frac{1}{x^2+1}dx -  \int \frac{x^2}{(x^2+1)^2}dx = $
$= \Biggr\rvert \begin{equation}
\begin{split}
& u = x \quad v' =\frac{x}{(x^2+1)^2} =\frac{1(x^2+1)'}{2(x^2+1)^2}\\
& u' = 1 \quad v = -\frac{1}{2} \frac{1}{x^2+1}
\end{split}
\end{equation} \Biggr\rvert$
$= \arctan x - (-x\frac{1}{2}\frac{1}{x^2+1} + \frac{1}{2} \int \frac{dx}{x^2+1})$
$= \arctan x + \frac{x}{2(x^2+1)} - \frac{1}{2}\arctan x + C = \frac{1}{2}\arctan x + \frac{x}{2(x^2+1)} + C$
Thank you.
 [/INST] So why not use the same trick? Since you know from lecture what $$f_2(x) = \int \frac{dx}{\left(x^2+1\right)^2}$$ is, you now have
$$
\begin{split}
f_3(x) 
 &= \int \frac{dx}{\left(x^2+1\right)^3}\\
 &= \int \frac{dx}{\left(x^2+1\right)^2} - \int x \cdot \frac{xdx}{\left(x^2+1\right)^3}\\
 &= f_2(x) - \text{proceed by parts as before}
\end{split}
$$
</s>