$\mathbb I =\int\limits_{0}^{1}(xe^{2x}-\frac{x}{\sqrt{4-x^2}})dx=\int\limits_{0}^{1} xe^{2x}dx - \int\limits_{0}^{1} \frac{x}{\sqrt{4-x^2}}dx =\mathbb J - \mathbb K.$ $\bigstar \mathbb J=\int\limits_{0}^{1} xe^{2x}dx$
Đặt $\begin{cases}u=x \\ dv=e^{2x}dx \end{cases},$ khi đó $\begin{cases}du=dx \\ v= \frac{1}{2}e^{2x} \end{cases},$ ta có:
$\mathbb J=\frac{1}{2}xe^{2x} \bigg|_0^1-\int\limits_{0}^{1}\frac{1}{2}e^{2x}dx=\frac{e^2}{2}-\frac{1}{4}e^{2x} \bigg|_0^1=\frac{e^2}{2}-\frac{1}{4}(e^2-1)=\frac{e^2}{4}+\frac{1}{4}$
$\bigstar \mathbb K=\int\limits_{0}^{1} \frac{x}{\sqrt{4-x^2}}dx $
Đặt $t=\sqrt{4-x^2}\Rightarrow t^2=4-x^2\Rightarrow 2tdt=-2xdx\Leftrightarrow tdt=-xdx,$ ta có:
$x=0\Rightarrow t=2;x=1\Rightarrow t=\sqrt 3$
$\mathbb K= \int\limits_{\sqrt 3}^{2} \frac{t}{t}dt=t \bigg|_{\sqrt 3}^2=2-{\sqrt 3}$
Suy ra: $\mathbb I= (\frac{e^2}{4}+\frac{1}{4})-(2-{\sqrt 3})=\frac{e^2}{4}+\sqrt 3 - \frac{7}{4}.$