A) \[{{e}^{\sqrt{x}}}+A\]
B) \[\frac{1}{2}{{e}^{\sqrt{x}}}+A\]
C) \[2(\sqrt{x}-1){{e}^{\sqrt{x}}}+A\]
D) \[2(\sqrt{x}+1){{e}^{\sqrt{x}}}+A\] (A is an arbitrary constant)
Correct Answer: C
Solution :
\[I=\int_{{}}^{{}}{{{e}^{\sqrt{x}}}.\,dx}\]. Put \[\sqrt{x}=t\Rightarrow \frac{1}{2\sqrt{x}}\,dx=dt\Rightarrow dx=2t\,dt\] \ \[I=\int_{{}}^{{}}{{{e}^{t}}.\,2t\,dt}=2\,[t\,.\,{{e}^{t}}-{{e}^{t}}]+A=2\,[\sqrt{x}\,.\,{{e}^{\sqrt{x}}}-{{e}^{\sqrt{x}}}]+A\] Þ \[I=2(\sqrt{x}-1)\,.\,{{e}^{\sqrt{x}}}+A\].
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