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${\log _e}(1 + x)$
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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