Since $$\left(\frac{x}{1+x^2}\right)'=\frac{2-(1+x^2)}{(1+x^2)^2}=\frac{2}{(1+x^2)^2}-\frac{1}{1+x^2}$$ one has $$2\int \frac{1}{(1+x^2)^2}dx=\frac{x}{1+x^2}+\int \frac{1}{1+x^2}dx$$ $$\Rightarrow \int\frac{1}{(1+x^2)^2}dx=\frac{x}{2(1+x^2)}+\frac{\tan^{-1} x}{2}+C$$

by using partial fraction $$\frac{1}{(1+x^2)^2}=\frac{1}{2}(\frac{1}{1+x^2}-\frac{x^2-1}{(x^2+1)^2})$$ $$=\frac{1}{2}(\frac{1}{1+x^2}-\frac{d}{dx}(\frac{x}{(x^2+1)}))$$

An idea that works for $\frac{1}{(1+x^2)^2}$ and for all other exponents $\frac{1}{(1+x^2)^n}$ is to get the base of the denominator in the numerator

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

The integral of the first term we know because it has a lower exponent and for the second we can write it as $$\frac{1}{2}x\frac{2x}{(1+x)^2}$$ and integrate by parts

$$\frac{1}{2}\int x\frac{2x}{(1+x^2)^2}=\frac{1}{2}x\frac{1}{(1+x^2)}+\frac{1}{2}\int\frac{1}{(1+x^2)}$$

which makes it boil down again to the integral with the smaller exponent, which we know.

Adding the two parts we get the answer $$\arctan(x)-\frac{1}{2}x\frac{1}{(1+x^2)}-\frac{1}{2}\arctan(x)=\frac{1}{2}\arctan(x)-\frac{1}{2}x\frac{1}{(1+x^2)}$$

Another way is to substitute $x=\tan(\theta)$. Then it boils down to $$\int \cos^2\theta d\theta = \theta/2 + \sin(2\theta)/2 + C$$

Now use $$\sin(2\theta) = 2\sin(\theta)\cos(\theta) = \frac{2\tan(\theta)}{1+\tan^2(\theta)}$$ Now put $\theta=\tan^{-1}x$ and you are done.

The easy approach: Add them to your integral table...

$$ \int{\frac{a}{{(a^2+x^2)}^2}dx=}\ \left(\frac{1}{{2a}^2}\right)arctan(x/a)+\left(\frac{1}{2a}\right)\frac{x}{a^2+x^2}\ +C$$

set a = 1:

$$\int{\frac{1}{{(1+x^2)}^2}dx}=\frac{1}{2}\left(arctan(x)+\frac{x}{1+x^2}\right)+C$$

If you need to prove this yourself, you can obtain the result above by taking the derivative of the following standard integral with respect a:

$$\int \frac{1}{a^{2}+x^{2}} dx = \frac{1}{a} arctan(x/a)$$