Question

$\int_{0}^{\infty}{\log\left( x + \frac{1}{x} \right)\frac{dx}{1 + x^{2}}}$ is equal to

• A. $\pi\log 2$
• B. $- \pi\log 2$
• C. $(\pi ⥂ / ⥂ 2)\log 2$
• D. $- (\pi ⥂ / ⥂ 2)\log 2$

Answer 1

Answer: (A)

$\pi\log 2$

Answer 2

$I = \int_{0}^{\infty}{\log\left( x + \frac{1}{x} \right)}\frac{1}{1 + x^{2}}dx$

Put $x = \tan\theta \Rightarrow dx = \sec^{2}\theta d\theta$

$\mathbf{\Rightarrow}\mathbf{I =}\int_{\mathbf{0}}^{\mathbf{\pi/2}}{\mathbf{\log}\mathbf{(}\mathbf{\tan}\mathbf{\theta}\mathbf{+}\mathbf{\cot}\mathbf{\theta}}\mathbf{)}\frac{\mathbf{\sec}^{\mathbf{2}}\mathbf{\theta}}{\mathbf{\sec}^{\mathbf{2}}\mathbf{\theta}}\mathbf{d\theta}$

⇒ $I = \int_{0}^{\pi/2}{\log(\tan\theta + \cot\theta})d\theta$

$\Rightarrow I = \int_{0}^{\pi/2}{\log\frac{(1 + \tan^{2}\theta)}{\tan\theta}d\theta}$

⇒ I $= 2\int_{0}^{\pi/2}{{logsec}\theta d\theta - \int_{0}^{\pi/2}{{logtan}\theta}}d\theta$

⇒ I $= 2\int_{0}^{\pi/2}{{logsec}\theta d\theta}$; $\left\{ \because\int_{0}^{\pi/2}{{logtan}\theta = 0} \right\}$

$\Rightarrow I = - 2\int_{0}^{\pi/2}{{logcos}\theta d\theta}$

⇒$I = - 2 \times \frac{- \pi}{2}\log 2$, $\left\{ \because\int_{0}^{\pi/2}{{logcos}\theta = - \frac{\pi}{2}\log 2} \right\}$

⇒ $I = \pi\log 2$.