The integrand is already a partial fraction, as pointed out by Qiaochu Yuan.
If we add and subtract $x^2$ in the numerator, we can integrate the first integral immediately
$$\begin{eqnarray*}
\int \frac{1}{\left( 1+x^{2}\right) ^{2}}dx &=&\int \frac{1}{1+x^{2}}dx-\int \frac{x^{2}}{\left( 1+x^{2}\right) ^{2}}dx
\\
&=&\arctan x-\int x\frac{x}{\left( 1+x^{2}\right) ^{2}}dx
\end{eqnarray*}$$
and the second integral by parts:
$$\begin{eqnarray*}
\int x\frac{x}{\left( 1+x^{2}\right) ^{2}}dx &=&x\left( -\frac{1}{2\left(
1+x^{2}\right) }\right) +\int \frac{1}{2\left( 1+x^{2}\right) }dx \\
&=&-\frac{x}{2\left( 1+x^{2}\right) }+\frac{1}{2}\arctan x.
\end{eqnarray*}$$
Added: by applying this method $n-1$ times, we can reduce the integration of the function $f(x)=\dfrac{1}{\left( 1+x^{2}\right) ^{n}}$ to the integration of $\dfrac{1}{1+x^{2}}.$