MCQ
$\int {\frac{{{e^{\sqrt x }}}}{{\sqrt x }}dx} = $
- A${e^{\sqrt x }}$
- B$\frac{{{e^{\sqrt x }}}}{2}$
- ✓$2\,{e^{\sqrt x }}$
- D$\sqrt x \,.\,{e^{\sqrt x }}$
Put $\sqrt x = t$, $\therefore \frac{1}{{2\sqrt x }}dx = dt$
$I = 2\int {{e^t}dt = 2{e^t} + C = 2{e^{\sqrt x }} + C} $.
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