How do I even begin calculating: $$\int^{\pi/4}_0 \frac{\tan^2x}{1+x^2}dx\text{ ?}$$I substituted $\tan^2x=\sec^2x-1$ but that did nothing. Then I did integration by parts to get $$\frac{\tan x-x}{1+x^2}+2\int_0^{\pi/4}\frac{x^2{\tan x-x}}{1+x^2}dx$$ which is harder than the original problem. Then I found this that seems to have some potential: $$\int_0^{\pi/4}\frac{\sec^2x-\tan^2x}{1+x^2}dx=\arctan\frac{\pi}{4}$$I have a feeling that this is the key, but I don't know what to do next. Also I tried to put the integral into a calculator which failed to calculate the exact answer (it approximated it to be $0.15$).
P.S. The problem can be found in page $4$ that has two infinite series on the top and bottom.
