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

 

Answers (1)

\\\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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