MCQ
$\int_{\,0}^{\,\infty } {\frac{{x\ln x\,dx}}{{{{(1 + {x^2})}^2}}}} $ is equal to
- ✓$0$
- B$1$
- C$\infty $
- DNone of these
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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$\mathrm{x}-2=-\mathrm{y}=\mathrm{z}-1,2(\mathrm{x}+1)=2(\mathrm{y}-1)=\mathrm{z}+1$
and be parallel to the line $\frac{x-2}{3}=\frac{y-1}{1}=\frac{z-2}{2}$.
Then which of the following points lies on $\mathrm{L}$ ?