Mathematics

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


ANSWER

$$0$$


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
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