MCQ
$\int\limits_0^{\sqrt 3 } {\,\,\frac{1}{2}\,} \,\frac{d}{{dx}}\,\left( {{{\tan }^{ - 1}}\frac{{2x}}{{1 - {x^2}}}} \right)dx$ equals
  • $\frac{\pi }{3}\,$
  • B
    $ - \,\,\frac{\pi }{6}\,$
  • C
    $\,\frac{\pi }{2}\,$
  • D
    $\,\frac{\pi }{4}\,$

Answer

Correct option: A.
$\frac{\pi }{3}\,$
a
$ta{n^{ - 1}}\left( {\frac{{2x}}{{1 - {x^2}}}} \right)=$$\left[ {\begin{array}{*{20}{c}}
  {2{{\tan }^{ - 1}}x\,\,\,\,\,\,\,\,\,if{\mkern 1mu} {\mkern 1mu} 0 < x < 1} \\ 
  {2{{\tan }^{ - 1}}x - \pi \,\,\,\,\,if{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} x > 1} 
\end{array}} \right.$

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