Q98 / 180 · COMEDK 2013 · Mathematics

$\int \sqrt{x} e^{\sqrt{x}} d x$ is equal to

  • A
    $2 \sqrt{x}-e^{\sqrt{x}}-4 \sqrt{x e^{\sqrt{x}}}+c$
  • Correct answer
    B
    $(2 x-4 \sqrt{x}+4) e^{\sqrt{x}}+c$
  • C
    $(2 x+4 \sqrt{x}+4) e^{\sqrt{x}}+c$
  • D
    $(1-4 \sqrt{x}) e^{\sqrt{x}}+c$

Solution

Let $I=\int \sqrt{x} e^{\sqrt{x}} d x$ Put $\sqrt{x}=t \Rightarrow x=t^{2} \Rightarrow d x=2 t d t$ $$ \begin{aligned} \therefore I &=2 \int t^{2} e^{t} d t \\ &=2\left[t^{2} e^{t}-\int e^{t} \cdot 2 t d t\right] \\ &=2\left[t^{2} e^{t}-2 \int t e^{t} d t\right] \\ &=2\left[t^{2} e^{t}-2\left(t e^{t}-\int e^{t} \cdot 1 d t\right)\right] \\ &=2\left[t^{2} e^{t}-2 t e^{t}+2 e^{t}\right]+c \\ &=\left(2 t^{2}-4 t+4\right) e^{t}+c \\ &=(2 x-4 \sqrt{x}+4) e^{\sqrt{x}}+c \end{aligned} $$

Hence, the correct answer is option: 'B'.

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