The integral $\int_{0}^{1} an ^{-1}\left(1-x+x^{2} ight) d x=log(P)$, Find P?

Question

The integral $\int_{0}^{1} 	an ^{-1}\left(1-x+x^{2}ight) d x=log(P)$, Find P?

Professor PC Thomas Classes · Answer

$\int_{0}^{1} 	an ^{-1}\left(\frac{1}{1-x+x^{2}}ight) d x$\ $=\int_{0}^{1} 	an ^{-1}\left(\frac{(1-x)+x}{1-x(1-x)}ight) d x$\ $=\int_{0}^{1}\left[	an ^{-1}(1-x)+	an ^{-1} xight] d x$\ $=\int_{0}^{1} 	an ^{-1}(1-x) d x+\int_{0}^{1} 	an ^{-1} x d x$\ $=\int_{0}^{1} 	an ^{-1}\{1-(1-x)\} d x+\int_{0}^{1} 	an ^{-1} x d x$\ $=2 \int_{0}^{1} 	an ^{-1} x d x\left[\int_{0}^{a} f(x) d x=\int_{0}^{a} f(a-x) d xight]$\ Now, $\int_{0}^{1} 	an ^{-1}\left(\frac{1}{1-x+x^{2}}ight) d x$\ $=\int_{0}^{1}\left[\frac{\pi}{2}-\cot ^{-1}\left(\frac{1}{1-x+x^{2}}ight)ight] d x$\ $=\frac{\pi}{2}-\int_{0}^{1} 	an ^{-1}\left(1-x+x^{2}ight) d x$\ $	herefore \int_{0}^{1} 	an ^{-1}\left(1-x+x^{2}ight) d x=\frac{\pi}{2}-2 I_{1}$\ Where, $I_{1}=\int_{0}^{1} 	an ^{-1} x d x$\ $=\left[x 	an ^{-1} xight]_{0}^{1}-\int_{0}^{1} \frac{x d x}{1+x^{2}}$\ $=\frac{\pi}{4}-\frac{1}{2}\left[\log \left(1+x^{2}ight)ight]_{0}^{1}=\frac{\pi}{4}-\frac{1}{2} \log 2$\ $	herefore \int_{0}^{1} 	an ^{-1}\left(1-x+x^{2}ight) d x=\log 2$\