MCQ
$\int_{\,0}^{\,\infty } {\frac{{x\ln x\,dx}}{{{{(1 + {x^2})}^2}}}} $ is equal to
  • $0$
  • B
    $1$
  • C
    $\infty $
  • D
    None of these

Answer

Correct option: A.
$0$
a
(a) $I = \int_0^\infty {\frac{{x\log x}}{{{{(1 + {x^2})}^2}}}\,dx} $

Put $x = \tan \theta $

==> $dx = {\sec ^2}\theta \,d\theta $

$\therefore I$ $ = \int_0^{\pi /2} {\frac{{\tan \theta \,\log \,(\tan \theta )}}{{{{\sec }^4}\theta }}} {\sec ^2}\theta \,d\theta $

$ = \int_0^{\pi /2} {\sin \theta \,\cos \theta \,\log \,(\tan \theta )\,d\theta } $

$ = \frac{1}{2}\int_0^{\pi /2} {\sin 2\theta \log \,(\,\tan \theta \,)\,d\theta } $$ = 0$,

$\left\{ \because \int_{0}^{\pi /2}{\sin 2\theta \,\,\log \,\,\tan \theta \,\,d\theta =0} \right\}$.

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