First slide

Evaluation of definite integrals

Question

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

Easy

A

0
B

1
C

$\sqrt{3} / 2$
D

$\sqrt{5} / 2$

Solution

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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