# The value of the integral $\int_{0}^{1}x\cot^{-1}(1-x^{2}+x^{4})dx$ is : Option 1) $\frac{\pi}{2}-\frac{1}{2}\:log_{e}\:2$ Option 2) $\frac{\pi}{4}-\:log_{e}\:2$ Option 3) $\frac{\pi}{2}-\:log_{e}\:2$ Option 4) $\frac{\pi}{4}-\frac{1}{2}\:log_{e}\:2$

$\\\int_{0}^{1}x\cot^{-1}(1-x^{2}+x^{4})dx\\\\\\\:= \int_{0}^{1}x\:\tan^{-1}\left ( \frac{1}{1-x^{2}+x^{4}} \right )dx$

$put\:x^{2}=t$

$\\\frac{1}{2}\:\int_{0}^{1} \:\tan^{-1}\left ( \frac{1}{1-t+t^{2}} \right )dt\\\\\\\:=\int_{0}^{1}\frac{1}{2}\tan^{-1}\left ( \frac{t+(1+t)}{1-t(1-t)} \right )dt$

$\\\frac{1}{2}\:\int_{0}^{1} \:(tan^{-1}t+\tan^{-1}(1-t))dt\\\\\\:=\frac{1}{2}\int_{0}^{1}\tan^{-1}(1-t)dt+\frac{1}{2}\int_{0}^{1}\tan^{-1}(1-t)dt$

$\\\int_{0}^{1} \:\tan^{-1}t\:dt=\int_{0}^{1}\tan^{-1}(1-t)dt$

$put \:\:\tan^{-1}t=k$

$\\\int_{0}^{\pi/4}k\:\sec^{2}k\:dk\\\\\\\:=\pi/4-1/2\:ln \:\:2\:\:( using\:\:\:by\:\:parts)$

Option 1)

$\frac{\pi}{2}-\frac{1}{2}\:log_{e}\:2$

Option 2)

$\frac{\pi}{4}-\:log_{e}\:2$

Option 3)

$\frac{\pi}{2}-\:log_{e}\:2$

Option 4)

$\frac{\pi}{4}-\frac{1}{2}\:log_{e}\:2$

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