MCQ
$\int_{}^{} {{e^{\sqrt x }}\;dx} $ is equal to
($A $ is an arbitrary constant)
- A${e^{\sqrt x }} + A$
- B$\frac{1}{2}{e^{\sqrt x }} + A$
- ✓$2(\sqrt x - 1){e^{\sqrt x }} + A$
- D$2(\sqrt x + 1){e^{\sqrt x }} + A$
($A $ is an arbitrary constant)
Put $\sqrt x = t \Rightarrow \frac{1}{{2\sqrt x }}\,dx = dt \Rightarrow dx = 2t\,dt$
$I = \int_{}^{} {{e^t}.\,2t\,dt} = 2\,[t\,.\,{e^t} - {e^t}] + A = 2\,[\sqrt x \,.\,{e^{\sqrt x }} - {e^{\sqrt x }}] + A$
==> $I = 2(\sqrt x - 1)\,.\,{e^{\sqrt x }} + A$.
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