^ _0 (x+1 x )1 1+x^2 dx=

MCQ

$\int\limits^{\infty}_0\log\Big(\text{x}+\frac{1}{\text{x}}\Big)\frac{1}{1+\text{x}^2}\text{ dx}=$

✓
$\pi\ln 2$
B
$-\pi\ln2$
C
$0$
D
$-\frac{\pi}{2}\ln2$

✓

Answer

Correct option: A.

$\pi\ln 2$

$\int\limits^{\infty}_0\log\Big(\text{x}+\frac{1}{\text{x}}\Big)\frac{1}{1+\text{x}^2}\text{ dx}$
Substitute $\text{x}=\tan\theta$
$\text{dx}=\sec^2\theta\text{d}\theta$
When,
$\text{x}=0\Rightarrow\theta=0$
$\text{x}=\infty\Rightarrow\theta=\frac{\pi}{2}$
$\int\limits^{\frac{\pi}{2}}_0\Big(\tan\theta+\frac{1}{\tan\theta}\Big)\frac{1}{1+\tan^{2}}\times\sec^2\theta\text{d}\theta$
$\int\limits^{\frac{\pi}{2}}_0\log\Big(\frac{\tan^2\theta+1}{\tan\theta}\Big)\frac{1}{1+\tan^{2}\theta}\times\sec^2\theta\text{d}\theta$
$\Rightarrow\int\limits^{\frac{\pi}{2}}_0\log\Big(\frac{\sec^2\theta}{\tan\theta}\Big)\frac{1}{\sec^2\theta}\times\sec^2\theta\text{d}\theta$ $\big[\because1+\tan^2\theta=\sec^2\theta\big]$
$\Rightarrow\int\limits^{\frac{\pi}{2}}_0\log\Big(\frac{\sec^2\theta}{\tan\theta}\Big)\text{d}\theta$
$\Rightarrow\int\limits^{\frac{\pi}{2}}_0\log\Big(\frac{1}{\sin\theta\cdot\cos\theta}\Big)\text{d}\theta$
$\Rightarrow\int\limits^{\frac{\pi}{2}}_0\log\big(\sin\theta\cos\theta\big)\text{d}\theta$
$\Rightarrow-\int\limits^{\frac{\pi}{2}}_0\big[\log\sin\theta+\log\cos\theta\big]\text{d}\theta$
$\Rightarrow-\int\limits^{\frac{\pi}{2}}_0\log\sin\theta\text{d}\theta-\int\limits^{\frac{\pi}{2}}_0\log\cos\theta\text{d}\theta$
Let us conside,
$\int\limits^{\frac{\pi}{2}}_0\log\sin\theta\text{d}\theta=\text{I}\ ....(\text{i})$
$\Rightarrow\text{I}=\int\limits^{\frac{\pi}{2}}_0\log\Big(\sin\Big(\frac{\pi}{2}-\theta\Big)\Big)\text{d}\theta$
$\Rightarrow\int\limits^{\frac{\pi}{2}}_0\log\cos\theta\text{d}\theta\ ...(\text{ii})$
Adding $(i)$ and $(ii)$
$2\text{I}=\int\limits^{\frac{\pi}{2}}_0\log\sin\theta\text{ d}\theta+\int\limits^{\frac{\pi}{2}}_0\log\cos\theta\text{ d}\theta$
$=\int\limits^{\frac{\pi}{2}}_0\log(\sin\theta\cdot\cos\theta)\text{d}\theta$
$=\int\limits^{\frac{\pi}{2}}_0\log\big(\sin2\theta\big)\text{d}\theta-\int\limits^{\frac{\pi}{2}}_0\log2\text{ d}\theta$
Let us consider $2\theta=\text{t}$
$2\text{d}\theta=\text{dt}$
$2\text{I}=\frac{1}{2}\int\limits^{\pi}_{0}\log(\sin\text{t})\text{dt}-\frac{\pi}{2}\log2$
$2\text{I}=\frac{2}{2}\int\limits^{\pi}_{0}\log(\sin\text{t})\text{dt}-\frac{\pi}{2}\log2$ $\big[\because\sin\theta$ is positive in both $1^{st}$ and $2^{nd} $ quadrants$\big]$
$2\text{I}=\text{I}-\frac{\pi}{2}\log2$
$2\text{I}-\text{I}=-\frac{\pi}{2}\log2$
$\text{I}=-\frac{\pi}{2}\log2,$ where $\text{I}=\int\limits^{\frac{\pi}{2}}_0\log\sin\theta\text{ d}\theta$
Now,
$-\int\limits^{\frac{\pi}{2}}_0\log(\sin\theta)\text{ d}\theta-\int\limits^{\frac{\pi}{2}}_0\log\cos\theta\text{ d}\theta$
$-2\int\limits^{\frac{\pi}{2}}_0\log\sin\theta\text{ d}\theta=-2\times\text{I}$
$=-2\times-\frac{\pi}{2}\log2$ $\Big[\text{Where I}=-\frac{\pi}{2}\log2\Big]$
$=\pi\log2$

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