Let $\displaystyle I = \int_{0}^{\frac{\pi}{4}}\tan^n xdx\;,$ Then $\displaystyle I_{n}+I_{n+2} = \int_{0}^{\frac{\pi}{4}}\tan^n x\cdot \sec^2 xdx = \frac{1}{n+1}$
So we get $\displaystyle f(x) = \frac{1}{n+1}$
Now $\displaystyle \sum^{10}_{n=1}(n+1)f(n) = \sum^{10}_{n=1}\frac{n+1}{n+1} = \sum^{10}_{n=1}1 = 10$