Mathematics

Solve : $$\displaystyle \underset{0}{\overset{\infty}{\int}} \dfrac{x \, \ln \, x}{(1 + x^2)^2}$$


SOLUTION
$$\int \dfrac{x\ln x}{(1+x^2)^2}dx$$

Integration by parts

$$=-\dfrac{\ln x}{2(1+x^2)}-\int -\dfrac{1}{2x(1+x^2)}dx$$

$$=-\dfrac{\ln x}{2(1+x^2)}-\left ( -\dfrac{1}{2}\left ( \left ( \ln |x| \right )-\dfrac{1}{2}\left ( \ln |x^2+1| \right ) \right ) \right )$$

$$\therefore \int \dfrac{x\ln x}{(1+x^2)^2}dx=-\dfrac{\ln x}{2(1+x^2)}+\dfrac{1}{2}\left ( \left ( \ln |x| \right )-\dfrac{1}{2}\left ( \ln |x^2+1| \right ) \right)+C$$

$$\therefore \int_{0}^{\infty }\dfrac{x\ln x}{(1+x^2)^2}dx=0-0=0$$
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Subjective Medium Published on 17th 09, 2020
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