Mathematics

# The value of integral  $\displaystyle \int_{0}^{\infty }\frac{x\log x}{(1+x^2)^2} \: dx$ is

$0$

##### SOLUTION
$\displaystyle I=\int_{0}^{\infty }\frac{x\log xdx}{(1+x^2)^2}$
Let $x=\frac{1}{t}$
$\displaystyle I=\int_{0}^{\infty }\frac{(\frac{1}{t})\log (\frac{1}{t}) (-\frac{1}{t^2})dt}{(1+(\frac{1}{t^2}))^2}$
$\displaystyle I=-\int_{0}^{\infty }\frac{t\log t}{(1+t^2)^2}dt=-I$
or $I=0$

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Single Correct Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 86

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