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 :
Given that, \[I=\int{\sqrt{x}\,{{e}^{\sqrt{x}}}d\,x}\] Putting \[\sqrt{x}=t\,\,\,\Rightarrow \,\,\frac{1}{2\sqrt{x}}dx=dt\] \[\therefore \] \[I=2\int{\,{{t}^{2}}{{e}^{t}}\,dt=2\,[{{t}^{2}}{{e}^{t}}-(2t){{e}^{t}}+2{{e}^{t}}]+c}\] \[=(2x-4\sqrt{x}+4)\,{{e}^{\sqrt{x}}}+c\]
You need to login to perform this action.
You will be redirected in
3 sec
