MCQ
If $\int_{}^{} {\frac{1}{{(1 + x)\sqrt x }}\;dx = f(x) + A} $, where $ A$  is any arbitrary constant, then the function $f(x)$ is
  • A
    $2{\tan ^{ - 1}}x$
  • $2{\tan ^{ - 1}}\sqrt x $
  • C
    $2{\cot ^{ - 1}}\sqrt x $
  • D
    ${\log _e}(1 + x)$

Answer

Correct option: B.
$2{\tan ^{ - 1}}\sqrt x $
b
(b) $I = \int_{}^{} {\frac{{dx}}{{\sqrt x (1 + {{(\sqrt x )}^2})}}} $

Put $\sqrt x = t \Rightarrow \frac{1}{{2\sqrt x }}\,dx = dt$
$I = \int_{}^{} {\frac{{2\,dt}}{{1 + {t^2}}} = 2{{\tan }^{ - 1}}t + A} $
$\therefore \,\,\,I = 2{\tan ^{ - 1}}\sqrt x + A$; $\therefore \,\,\,f(x) = 2{\tan ^{ - 1}}\sqrt x $.

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