e^ ^ - 1 x x + x x dx = - Vidyadip

MCQ

$\int {\frac{{{e^{{{\tan }^{ - 1}}\sqrt x }}}}{{\sqrt x + x\sqrt x }}dx = } $

A
${e^{{{\tan }^{ - 1}}\sqrt x }} + c$
B
$\frac{1}{2}{e^{{{\tan }^{ - 1}}\sqrt x }} + c$
C
$\log {\tan ^{ - 1}}\sqrt x + c$
✓
$2{e^{{{\tan }^{ - 1}}\sqrt x }} + c$

✓

Answer

Correct option: D.

$2{e^{{{\tan }^{ - 1}}\sqrt x }} + c$

d
Put $\tan ^{-1} \sqrt{x}=t$

$\frac{1}{1+x} \times \frac{1}{2 \sqrt{x}} d x=d t$

$\frac{d x}{\sqrt{x}+x \sqrt{x}}=2 d t$

$\int \mathrm{e}^{\mathrm{t}} 2 \mathrm{dt}=2 \mathrm{e}^{\mathrm{t}}+\mathrm{c}$

$=2 \mathrm{e}^{\tan ^{-1} \sqrt{\mathrm{x}}}+\mathrm{c}$

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