Question

\[\int\limits_0^1 \frac{\log\left( 1 + x \right)}{1 + x^2} dx\]

 

Answer 1

We have,

\[I = \int\limits_0^1 \frac{\log \left( 1 + x \right)}{1 + x^2} dx\]

\[Putting\ x = \tan \theta\]

\[ \Rightarrow dx = \sec^2 \theta d\theta\]

\[\text{When }x \to 0 ; \theta \to 0\]

\[\text{and }x \to 1 ; \theta \to \frac{\pi}{4}\]

\[\text{Now, integral becomes}\]

\[I = \int\limits_0^\frac{\pi}{4} \frac{\log \left( 1 + \tan \theta \right)}{\sec^2 \theta} \sec^2 \theta d\theta\]
\[ \Rightarrow I = \int\limits_0^\frac{\pi}{4} \left[ \log \left( 1 + \tan \theta \right) \right] d\theta ................\left( 1 \right)\]
\[ \Rightarrow I = \int\limits_0^\frac{\pi}{4} \left[ \log\left\{ 1 + \tan \left( \frac{\pi}{4} - \theta \right) \right\} \right] d\theta ...................\left[ \because \int_0^a f\left( x \right)dx = \int_0^a f\left( a - x \right)dx \right]\]
\[ = \int\limits_0^\frac{\pi}{4} \left[ \log\left\{ 1 + \frac{\tan\frac{\pi}{4} - \tan \theta}{1 + \tan\frac{\pi}{4} \tan \theta} \right\} \right] d\theta\]
\[ = \int\limits_0^\frac{\pi}{4} \left[ \log\left\{ 1 + \frac{1 - \tan \theta}{1 + \tan \theta} \right\} \right] d\theta\]
\[ = \int\limits_0^\frac{\pi}{4} \left[ \log\left\{ \frac{2}{1 + \tan \theta} \right\} \right] d\theta\]
\[I = \int_0^\frac{\pi}{4} \left[ \log 2 - \log \left( 1 + \tan \theta \right) \right] d\theta . . . . . \left( 2 \right)\]

\[\text{Adding} \left( 1 \right) and \left( 2 \right), \text{we get}\]

\[2I = \int_0^\frac{\pi}{4} \left( \log 2 \right) d\theta\]

\[ \Rightarrow 2I = \left( \log 2 \right) \left[ \theta \right]_0^\frac{\pi}{4} \]

\[ \Rightarrow 2I = \frac{\pi}{4}\log 2\]

\[ \Rightarrow I = \frac{\pi}{8}\log 2\]

\[ \therefore \int\limits_0^1 \frac{\log\left( 1 + x \right)}{1 + x^2}dx = \frac{\pi}{8}\log 2\]