Question

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

Answer 1

We have,

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

\[\text{Putting }x = \tan \theta\]

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

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

\[\text{and }x \to \infty ; \theta \to \frac{\pi}{2}\]

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

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

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

\[2I = \int\limits_0^\frac{\pi}{2} \frac{\sin\theta + \cos\theta}{\sin\theta + \cos\theta} d\theta\]

\[ \Rightarrow 2I = \int\limits_0^\frac{\pi}{2} d\theta\]

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

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

\[ \therefore \int\limits_0^\infty \frac{x}{\left( 1 + x \right)\left( 1 + x^2 \right)} dx = \frac{\pi}{4}\]