MCQ
$\int_{}^{} {\frac{{dx}}{{2\sqrt x (1 + x)}} = } $
- A$\frac{1}{2}{\tan ^{ - 1}}(\sqrt x ) + c$
- ✓${\tan ^{ - 1}}(\sqrt x ) + c$
- C$2{\tan ^{ - 1}}(\sqrt x ) + c$
- DNone of these
Put $\sqrt x \, = t$==> $\frac{1}{{2\sqrt x }}dx = dt$
$\therefore I = \int {\frac{{dt}}{{1 + {t^2}}}} = {\tan ^{ - 1}}t + c$$ = {\tan ^{ - 1}}(\sqrt x ) + c$.
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$x = 1 + xy\frac{{dy}}{{dx}} + \frac{{{{\left( {xy} \right)}^2}}}{{2!}}{\left( {\frac{{dy}}{{dx}}} \right)^2} + \frac{{{{\left( {xy} \right)}^3}}}{{3!}}{\left( {\frac{{dy}}{{dx}}} \right)^3} + ......$ is