The integral 0 1 tan - 1 x 1 + x ? d x equals

Question

$The integral \int_{0}^{1} \frac{\tan^{- 1} x}{1 + x} d x equals$

Smart Achievers · Accepted Answer

(3)

$Let x = \tan θ$

$\int_{0}^{π / 4} \frac{θ}{1 + \tan θ} \sec^{2} θ d θ$

$= θ \log_{e} (1 + \tan θ) I_{0}^{π / 4} - \int_{0}^{π / 4} \log_{e} (1 + \tan θ) d θ$ $= \frac{π}{4} \log_{e} 2 - I$

$I = \int_{0}^{π / 4} \log_{e} (1 + \tan θ) d θ$ $. . . (1)$

$I = \int_{0}^{π / 4} \log_{e} (1 + \tan (\frac{π}{4} - θ)) d θ$

$I = \int_{0}^{π / 4} \log_{e} (\frac{2}{1 + \tan θ}) d θ$

$By (1) + (2) \Rightarrow 2 I = \int_{0}^{π / 4} \log_{e} 2 d θ = \frac{π}{4} \log_{e} 2$

$\Rightarrow I = \frac{π}{8} \log_{e} 2$

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