If 2\int_{0}^{1}\tan ^{-1}xdx = \int_{0}^{1}\cot ^{-1}(1-x+x^{2})dx , then \int_{0}^{1}\tan ^{-1}(1-x+x^{2})dx

is equal to :

  • Option 1)

    log4

  • Option 2)

    \frac{\pi }{2}+log 2

  • Option 3)

    log2

  • Option 4)

    \frac{\pi }{2}-log 4

 

Answers (1)

As learnt in concept

Integration By PARTS -

Let u and v are two functions then 

\int u\cdot vdx=u\int vdx-\int \left ( \frac{du}{dx}\int vdx \right )dx

- wherein

Where u is the Ist function v is he IInd function

 

 2\int_{0}^{1} \tan ^{-1}x dx=\int_{0}^{1}\cot ^{-1}\left ( 1-x+x^{2} \right )dx ,

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

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

\int_{0}^{1}\tan ^{-1}\left ( 1-x+x^{2} \right )dx

=\frac{\pi }{2}-2\left [ x\tan ^{-1} x-\frac{1}{2 }\:log\left ( 1+x^{2} \right )\right ]^{1}_{0}

=\frac{\pi }{2}-2\left [ \frac{\pi }{4}-\frac{1}{2}\: log \:2-0 \right ]

=log2

 


Option 1)

log4

Incorrect option

Option 2)

\frac{\pi }{2}+log 2

Incorrect option

Option 3)

log2

Correct option

Option 4)

\frac{\pi }{2}-log 4

Incorrect option

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