The value of ∫01/3tan−1x+cot−1x(tan−1x−cot−1x) dx1+x2is /are

π2∫01/3dx(2tan−1x−π4)(1+x2), π2∫01/3dx(tan−1x−π4)(1+x2)tan−1=y dx1+x2=dy=π4∫0π/6dyy−π4=π4log|y−π4|0π6=π4log|π/2π/4|=π4log13 = - π4log3

The value of ∫01/3tan−1x+cot−1x(tan−1x−cot−1x) dx1+x2is /are

π2∫01/3dx(2tan−1x−π4)(1+x2), π2∫01/3dx(tan−1x−π4)(1+x2)tan−1=y dx1+x2=dy=π4∫0π/6dyy−π4=π4log|y−π4|0π6=π4log|π/2π/4|=π4log13 = - π4log3

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The value of $\int_{0}^{1 / \sqrt{3}} \frac{\tan^{- 1} x + \cot^{- 1} x}{(\tan^{- 1} x - \cot^{- 1} x)} \frac{d x}{1 + x^{2}}$ is /are

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$- \frac{π}{4} \log 3^{}$

$\frac{π}{4} \log \frac{1}{3}$

$\frac{π}{4} \log 2$

$\frac{π}{8} \log 4$

answer is A, B.

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Detailed Solution

$\begin{array}{l} \frac{π}{2} \int_{0}^{1 / \sqrt{3}} \frac{d x}{(2 \tan^{- 1} x - \frac{π}{4}) (1 + x^{2})}, \frac{π}{2} \int_{0}^{1 / \sqrt{3}} \frac{d x}{(\tan^{- 1} x - \frac{π}{4}) (1 + x^{2})} \\ \tan^{- 1} = y \frac{d x}{1 + x^{2}} = d y = \frac{π}{4} \int_{0}^{π / 6} \frac{d y}{y - \frac{π}{4}} = \frac{π}{4} \log {| y - \frac{π}{4} |}_{0}^{\frac{π}{6}} \\ = \frac{π}{4} \log | \frac{π / 2}{π / 4} | = \frac{π}{4} \log^{\frac{1}{3}} = - \frac{π}{4} \log^{3} \end{array}$