Đặt $\sqrt{4+2x}=t$ thì $x=\frac{t^2-4}{2}$, suy ra $dx=tdt$.Ta có $\int\limits_{0}^{1}\frac{x^2dx}{\sqrt{4+2x}}=\int\limits_{2}^{\sqrt{6}}\frac{\frac{(t^2-4)^2}{4}.tdt}{t}=\frac{1}{4}\int\limits_{2}^{\sqrt{6}}(t^4-4t^2+16)dt=...$
Ta có $\int\limits_{0}^{1}{\frac{x^2dx}{\sqrt{4+2x}}}=\lim_{c\to 0}\int\limits_{c}^{1}{\frac{x^2dx}{\sqrt{4+2x}}}$Đặt $\sqrt{4+2x}=t$ thì $x=\frac{t^2-4}{2}$, suy ra $dx=tdt$.Ta có $\int\limits_{0}^{1}\frac{x^2dx}{\sqrt{4+2x}}=\int\limits_{2}^{\sqrt{6}}\frac{\frac{(t^2-4)^2}{4}.tdt}{t}=\frac{1}{4}\int\limits_{2}^{\sqrt{6}}(t^4-4t^2+16)dt=...$
Đặt $\sqrt{4+2x}=t$ thì $x=\frac{t^2-4}{2}$, suy ra $dx=tdt$.Ta có $\int\limits_{0}^{1}\frac{x^2dx}{\sqrt{4+2x}}=\int\limits_{2}^{\sqrt{6}}\frac{\frac{(t^2-4)^2}{4}.tdt}{t}=\frac{1}{4}\int\limits_{2}^{\sqrt{6}}(t^4-4t^2+16)dt=...$