Question

$\int \sqrt{x}{e}^{\sqrt{X}}dx=$

• 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. $(1+4\sqrt{x})e^{\sqrt{X}}+c$

Answer 1

Answer: (B)

$(2x-4\sqrt{x}+4)e^{\sqrt{X}}+c$

$\int \sqrt{x}{e}^{\sqrt{x}}dx$

$let\sqrt{x}=t$

$\frac{1}{2\sqrt{x}}dx=dt$

$dx=2tdt$

$\int t{e}^t\left(2tdt\right)$

$=2\int t^2{e}^{t}dt$

$\begin{array}{c}\int t^2{e}^{t}dt=t^2{e}^t-2\int t{e}^{t}dt\\ \int t{e}^{t}dt=t{e}^t-e^t\end{array}\}\int t^2{e}^{t}dt$

$=t^2{e}^t-[t{e}^t-e^t]+c$

$=t^2{e}^t-t{e}^t+e^t+c$

$=e^t[t^2-2t+2]+c\cdot$

So, $2\int t^2{e}^{t}dt=e^t(2t^2-4t+4)+c$

$e^{\sqrt{x}}[2x-4\sqrt{x}+4]+c.$