The value of ∫0∞ xlog⁡x1+x22dx is

Putting u=1/xI=∫0∞ xlog⁡x1+x2dx=∫∞0 ulog⁡u1+u22du=−ISo I =0.

The value of ∫0∞ xlog⁡x1+x22dx is

Putting u=1/xI=∫0∞ xlog⁡x1+x2dx=∫∞0 ulog⁡u1+u22du=−ISo I =0.

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

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Putting $u = 1 / x$

$I = \int_{0}^{\infty} \frac{x \log x}{(1 + x^{2})} d x = \int_{\infty}^{0} \frac{u \log u}{{(1 + u^{2})}^{2}} d u = - I$

So $I = 0 .$

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