0-x-log-x-1-x-2-2-dx-

Question Number 59049 by ugwu Kingsley last updated on 03/May/19

\underset{0}{\int^{\infty}} \frac{x \log x}{{(1 + x^{2})}^{2}} d x =

Commented by maxmathsup by imad last updated on 04/May/19

l e t A = \int_{0}^{\infty} \frac{x l n (x)}{{(1 + x^{2})}^{2}} d x \Rightarrow A = \int_{0}^{1} \frac{x l n (x)}{{(1 + x^{2})}^{2}} d x + \int_{1}^{+ \infty} \frac{x l n (x)}{{(1 + x^{2})}^{2}} d x

\int_{1}^{+ \infty} \frac{x l n (x)}{{(1 + x^{2})}^{2}} d x =_{x = \frac{1}{t}} - \int_{0}^{1} \frac{1}{t {(1 + \frac{1}{t^{2}})}^{2}} l n (\frac{1}{t}) \frac{- d t}{t^{2}}

= - \int_{0}^{1} \frac{l n (t)}{t^{3} {(\frac{1 + t^{2}}{t^{2}})}^{2}} d t = - \int_{0}^{1} \frac{t l n (t)}{{(1 + t^{2})}^{2}} \Rightarrow A = 0 .