Paneer Definite Masala 7

Putting $y = x - x^{-1}$ , this integral becomes $I \; = \; \int_0^1 \frac{\ln(1+y)}{y^2+1}\,dy$ Putting $y = \tan\theta$ , we obtain $I \;=\; \int_0^{\frac14\pi}\ln(1 + \tan\theta)\,d\theta$ Putting $\theta = \tfrac14\pi - \phi$ we obtain $I \; = \; \int_0^{\frac14\pi} \ln\left(\frac{2}{1 + \tan\phi}\right)\,d\phi \; = \; \tfrac14\pi\ln2 - I \; =\; \tfrac18\pi\ln2$ making the answer $1 + 1 + 2\times8 + 2 = \boxed{20}$ .