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) \[(1+4\sqrt{x}){{e}^{\sqrt{x}}}+c\]
D) None of the above
Correct Answer: B
Solution :
Let \[I=\int{{{\sqrt{xe}}^{\sqrt{x}}}dx}\] Put \[\sqrt{x}=t\] \[\Rightarrow \] \[\frac{1}{2\sqrt{x}}dx=dt\] \[\Rightarrow \] \[dx=2t\,dt\] \[\therefore \] \[I=\int{t{{e}^{t}}.2t\,dt}\] \[\Rightarrow \] \[I=2\int{{{t}^{2}}.{{e}^{t}}}dt\] \[\Rightarrow \] \[I=2[{{e}^{t}}.{{t}^{2}}-\int{2t{{e}^{t}}dt}]\] \[\Rightarrow \] \[I=2[{{e}^{t}}{{t}^{2}}-2\{t{{e}^{t}}-{{e}^{t}}\}]+c\] \[\Rightarrow \] \[I=2{{e}^{t}}{{t}^{2}}-4t{{e}^{t}}+4{{e}^{t}}+c\] \[\Rightarrow \] \[I={{e}^{t}}(2{{t}^{2}}-4t+4)+c\] \[\Rightarrow \] \[I={{e}^{\sqrt{x}}}(2x-4\sqrt{x}+4)+c\]
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