Mathematics

# $\displaystyle \int_{0}^{\infty}\frac{\log(1+x^{2})}{1+x^{2}}dx=$

$\pi$ log2

##### SOLUTION
Let $I = \displaystyle\int_{0}^{\infty }\dfrac{\log(1+x^{2})}{1+x^{2}}dx$
Substitute  $x=\tan\theta$
Then, $I=\displaystyle\int_{0}^{\pi/2}-2\log(\cos\theta )d\theta$
Let $L=\displaystyle\int_{0}^{\pi/2}\log(\cos\theta )d\theta$       .....(1)
Using $(a+b-x)$ rule,
$L=\displaystyle\int_{0}^{\pi/2}\log(\sin\theta)d\theta$           .....(2)
Adding (1) and (2), we get
$L=\displaystyle\int_{0}^{\pi/2}\dfrac{1}{2}\log(\sin\theta \text{ }\cos\theta) d\theta$
$L=\dfrac{1}{2}\displaystyle\int_{0}^{\dfrac{\pi }{2}}\log(\sin 2\theta )d\theta -\dfrac{\pi }{4}\log(2)$
Substituting $2\theta =\dfrac{\pi }{2}-t$, we get
$L=\dfrac{1}{4}\displaystyle \int_{-\pi/2}^{\pi/2}\log(\cos t)dt-\dfrac{\pi}{4}\log(2)=\dfrac{1}{2}\displaystyle\int_{0}^{\pi/2}\log(\cos t)dt-\dfrac{\pi}{4}\log(2)$
$\Rightarrow L=\dfrac{L}{2} -\dfrac{\pi }{4}\log(2)$
$\Rightarrow L=-\dfrac{\pi }{2}\log(2)$
Since, $I=-2L$
$\Rightarrow I=\pi \log(2)$
So, option A is correct.

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Single Correct Hard Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 111

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