Integrate squared trigonometric function I'm trying to integrate $\int_a^b \left( \frac{1}{1+x^2}  \right)^2 dx$
I know that $\frac{d}{dx} \arctan(x) = \frac{1}{1+x^2}$, but how can I integrate with the squared part?
I've tried substitution with no success.

Full Work
Perform a substitution: 
$$x = tan(u)$$
$$dx = sec^2(u)du$$
Then:
$$\int\nolimits\frac{sec^2(u)}{(1+tan^2(u))^2} du$$
Use the trig identity:
$$1+tan^2(x) = sec^2(x)$$
Then:
$$\int\nolimits\frac{sec^2(u)}{sec^4(u)} du$$
Simplify:
$$\int\nolimits\frac{1}{sec^2(u)} du$$
$$\int\nolimits{cos^2(u)} du$$
Use the identity:
$$\cos^{2}(x) = \frac{1}{2}(1+\cos(2x))$$
Then:
$$\frac{1}{2}\int\nolimits{(1+\cos(2u))} du$$
Which equals:
$$\frac{1}{2}u + \frac{1}{4}\sin(2u) du$$
Now either sub back:
$$ x = tan(u)$$ 
Or solve for new limits, by plugging a & b in to:
$$ x = tan(u)$$
