\(I=\int_{0}^{1}x^{2}(1+x\sqrt{1-x^{2}})dx=\int_{0}^{1}x^{2}dx+\int_{0}^{1}x^{3}\sqrt{1-x^{2}}dx\)
\(I_{1}=\int_{0}^{1}x^{2}dx=\frac{x^{3}}{3} \bigg | _{0}^{1}=\frac{1}{3}\)
\(I_{2}=\int_{0}^{1}x^{3}\sqrt{1-x^{2}}dx\)
Đặt \(t=\sqrt{1-x^{2}}\Rightarrow x^{2}=1-t^{2}\Rightarrow xdx=-tdt\)
Đổi cận: \(x=0\Rightarrow t=1;x=1\Rightarrow t=0\)
\(\Rightarrow I_{2}=-\int_{1}^{0}(1-t^{2})t^{2}dt=\int_{0}^{1}(t^{2}-t^{4})dt=\left ( \frac{t^{3}}{3}-\frac{t^{5}}{5} \right ) \bigg |_{0}^{1}=\frac{2}{15}\)
Vậy \(I=I_{1}+I_{2}=\frac{7}{15}\)
Đặt \(u=x\Rightarrow du=dx;dv=e^{2x}dx\) chọn \(v=\frac{1}{2}e^{2x}\)
\(\Rightarrow \int_{0}^{1}xe^{2x}dx=\frac{x}{2}e^{2x}|_{0}^{1}-\frac{1}{2}\int_{0}^{1}e^{2x}dx=\frac{e^{2}}{2}-\frac{1}{4}e^{2x}|_{0}^{1}=\frac{e^{2}+1}{4}\)
Vậy \(I=\frac{3e^{2}+7}{12}.\)
