QJEEM11Ma8b0c: JEE Main 2011 Mathematics - Class 12 JEEM11Ma8b0c Solution

Question

Verified Solution • 2025–26 Session moderate

The value of

\int\limits_0^1 {{{8\log \left( {1 + x} \right)} \over {1 + {x^2}}}} dx

is

Accepted Answer

I = \int\limits_0^1 {{{8\log \left( {1 + x} \right)} \over {1 + {x^2}}}} dx

put

x = \tan \,\theta ,

\therefore

{{dx} \over {d\theta }} = {\sec ^2}\theta \Rightarrow dx = {\sec ^2}\theta d\theta

\therefore

I = 8\int\limits_0^{\pi /4} {{{\log \left( {1 + \tan \theta } \right)} \over {1 + {{\tan }^2}\theta }}} .{\sec ^2}\theta d\theta

I = 8\int\limits_0^{\pi /4} {\log \left( {1 + \tan \theta } \right)} d\theta \,...\left( i \right)

= 8\int\limits_0^{\pi /4} {\log \left[ {1 + \tan \left( {{\pi \over 4} - \theta } \right)} \right]} d\theta

= 8\int\limits_0^{\pi /4} {\log \left[ {1 + {{1 - \tan \theta } \over {1 + \tan \theta }}} \right]} d\theta

= 8\int\limits_0^{\pi /4} {\log \left[ {{2 \over {1 + \tan \theta }}} \right]} d\theta

= 8\int\limits_0^{\pi /4} {\left[ {\log 2 - \log \left( {1 + \tan \theta } \right)} \right]d\theta }

I = 8.\left( {\log 2} \right)\left[ x \right]_0^{\pi /4} - 8

\,\,\,\,\,\,\,\,\,\int\limits_0^{\pi /4} {\log \left( {1 + \tan \theta } \right)} d\theta

I = 8.{\pi \over 4}.\log 2 - I

\left[ {} \right.

From equation

(i)

\left. {} \right]

\Rightarrow 2I = 2\pi \log 2,

\therefore

I = \pi \log 2

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