MCQ
$\int_{\,0}^{\,\infty } {\frac{{xdx}}{{(1 + x)(1 + {x^2})}} = } $
- A$0$
- B$\pi /2$
- ✓$\pi /4$
- D$1$
$ = \left[ {\frac{{ - 1}}{2}\log (1 + x)} \right]_0^\infty + \frac{1}{2} \times \frac{1}{2}[\log \,(1 + {x^2})]\,_0^\infty + \frac{1}{2}[{\tan ^{ - 1}}x]\,_0^\infty $
$ = 0 + 0 + \frac{1}{2}\left[ {\frac{\pi }{2} - 0} \right] = \frac{\pi }{4}$.
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$f(x) = \left\{ {\begin{array}{*{20}{c}}
{{x^2} + 2mx - 1\,,}&{x \leq 0}\\
{mx - 1\,\,\,\,\,\,\,\,\,\,\,\,\,,}&{x > 0}
\end{array}} \right.$
જો $f (x)$ એક-એક વિધેય હોય તો $'m'$ ની કિમતોનો ગણ મેળવો.