The integral dx equals

JEE MainMathematicsModerateDefinite Integration

The integral $\int_{0}^{1}\mspace{2mu}\frac{\tan^{- 1}x}{1 + x}$
dx equals

Show answer and solution

Let x =

\tan\theta

\int_{0}^{\pi/4}\mspace{2mu}\frac{\theta}{1 + tan\theta}\sec^{2}\theta d\theta = \theta\log_{e}

\left. \ (1 + tan\theta) \right|_{0}^{\pi/4} - \int_{0}^{\pi/4}\mspace{2mu}\log_{e}(1 + tan\theta)d\theta = \frac{\pi}{4}\log_{e}2 - I

I = \int_{0}^{\pi/4}\mspace{2mu}\log_{e}(1 + tan\theta)d\theta

I = \int_{0}^{\pi/4}\mspace{2mu}\log_{e}\left( 1 + tan\left( \frac{\pi}{4} - \theta \right) \right)d\theta = \int_{0}^{\pi/4}\mspace{2mu}\log_{e}\left( \frac{2}{1 + tan\theta} \right)d\theta

2I = \int_{0}^{\pi/4}\mspace{2mu}\log_{e}2d\theta = \frac{\pi}{4}\log_{e}2

\Rightarrow

I = \frac{\pi}{8}\log_{e}2

Source: Grass-G / M-13 Mathematics Paper 4 Nov.docx

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