If $\int\left(\frac{1}{1-x^8} ight)\left[\cos ^{-1}\left(\frac{2 x}{1+x^2} ight)+ an ^{-1}\left(\

Question

If $\int\left(\frac{1}{1-x^8}ight)\left[\cos ^{-1}\left(\frac{2 x}{1+x^2}ight)+	an ^{-1}\left(\frac{2 x}{1-x^2}ight)ight] d x$. has $p 	an ^{-1} f(x) \& q 	an ^{-1} g(x)$ terms and $x \in(-1,1)$, then: (where $p$ and $q$ are constant)

Professor PC Thomas Classes · Answer

$\int(\frac{1}{1-x^{8}})[cos^{-1}(\frac{2x}{1+x^{2}})+tan^{-1}(\frac{2x}{1-x^{2}})]dx$ $=\int(\frac{1}{1-x^{8}})[\frac{\pi}{2}-sin^{-1}\frac{2x}{1+x^{2}}+tan^{-1}(\frac{2x}{1-x^{2}})]dx$ $=\frac{\pi}{8}\iint(\frac{1}{1+x^{2}}+\frac{1}{1-x^{2}})dx+\frac{\pi}{4}\iint\frac{dx}{x^{4}+1}=\frac{\pi}{8}(tan^{-1}x+\frac{1}{2}ln|\frac{1+x}{1-x}|)+\frac{\pi}{8}\iint[\frac{x^{2}+1}{x^{4}+1}-\frac{x^{2}-1}{x^{4}+1}]dx$ $=\frac{\pi}{8}|tan^{-1}x+\frac{1}{2}ln|\frac{1+x}{1-x}|+\frac{1}{\sqrt{2}}tan^{-1}\frac{x^{2}-1}{x\sqrt{2}}-\frac{1}{2\sqrt{2}}ln|\frac{x^{2}-x\sqrt{2}+1}{x^{2}+x\sqrt{2}}||+c$