The value of ∫−1313cos−1(2x1+x2)+Tan−1(2x1−x2)ex+1dx is 4πk3. Then k=

I=∫−1313cos−1(2x1+x2)+Sin−1(2x1+x2)ex+1dx⇒I=π2.∫−13131ex+1dx⇒I=π2.∫0131ex+1+1e-x+1dxI=π2∫0131dx=π23

The value of ∫−1313cos−1(2x1+x2)+Tan−1(2x1−x2)ex+1dx is 4πk3. Then k=

I=∫−1313cos−1(2x1+x2)+Sin−1(2x1+x2)ex+1dx⇒I=π2.∫−13131ex+1dx⇒I=π2.∫0131ex+1+1e-x+1dxI=π2∫0131dx=π23

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The value of $\int_{\frac{- 1}{\sqrt{3}}}^{\frac{1}{\sqrt{3}}} \frac{\cos^{- 1} (\frac{2 x}{1 + x^{2}}) + T a n^{- 1} (\frac{2 x}{1 - x^{2}})}{e^{x} + 1} d x$ is $\frac{4 π}{k \sqrt{3}} .$ Then $k =$

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$I = \int_{\frac{- 1}{\sqrt{3}}}^{\frac{1}{\sqrt{3}}} \frac{\cos^{- 1} (\frac{2 x}{1 + x^{2}}) + S i n^{- 1} (\frac{2 x}{1 + x^{2}})}{e^{x} + 1} d x$

$\Rightarrow I = \frac{π}{2} . \int_{\frac{- 1}{\sqrt{3}}}^{\frac{1}{\sqrt{3}}} \frac{1}{e^{x} + 1} d x$

$\Rightarrow I = \frac{π}{2} . (\int_{0}^{\frac{1}{\sqrt{3}}} \frac{1}{e^{x} + 1} + \frac{1}{e^{- x} + 1}) d x$

$I = \frac{π}{2} \int_{0}^{\frac{1}{\sqrt{3}}} 1 d x = \frac{π}{2 \sqrt{3}}$

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The value of ∫−1313cos−1(2x1+x2)+Tan−1(2x1−x2)ex+1dx is 4πk3. Then k=

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The value of $\int_{\frac{- 1}{\sqrt{3}}}^{\frac{1}{\sqrt{3}}} \frac{\cos^{- 1} (\frac{2 x}{1 + x^{2}}) + T a n^{- 1} (\frac{2 x}{1 - x^{2}})}{e^{x} + 1} d x$ is $\frac{4 π}{k \sqrt{3}} .$ Then $k =$