A) \[2\sqrt{x}-{{e}^{\sqrt{x}}}-4\sqrt{x}\ {{e}^{\sqrt{x}}}+c\]
B) \[(2x-4\sqrt{x}+4){{e}^{\sqrt{x}}}+c\]
C) \[(2x+4\sqrt{x}+4){{e}^{\sqrt{x}}}+c\]
D) \[(1-4\sqrt{x}){{e}^{\sqrt{x}}}+c\]
Correct Answer: B
Solution :
\[I=\int_{{}}^{{}}{\sqrt{x}.{{e}^{\sqrt{x}}}}dx\]. Let \[x={{t}^{2}}\Rightarrow dx=2t\,dt\] \[\therefore I=2\int_{{}}^{{}}{{{t}^{2}}}\,.\,{{e}^{t}}dt\] Þ \[I=2({{t}^{2}}.{{e}^{t}}-2t{{e}^{t}}+2{{e}^{t}}]+c\] Þ \[I=\frac{1}{\sqrt{2}}\log \left| \tan \left( \frac{x}{2}+\frac{3\pi }{8} \right)\, \right|+c\] i.e., \[I={{e}^{\sqrt{x}}}[2x-4\sqrt{x}+4]+c\].
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