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 $\dfrac{1}{1+x^{2}}.$