Ta có \(I=\int_{1}^{2}x(\sqrt{x+1}-lnx)dx=\int_{1}^{2}x\sqrt{x+1}dx-\int_{1}^{2}xlnxdx=I_1+I_2\) Tính \(I_1=\int_{1}^{2}x\sqrt{x+1}dx\) Đặt \(t=\sqrt{x+1}\Rightarrow t^2=x+1\Rightarrow 2tdt=dx\) Đổi cận: \(x=2\Rightarrow t=\sqrt{3}\) \(x=1\Rightarrow t=\sqrt{2}\) Vậy \(I_1=\int_{\sqrt{2}}^{\sqrt{3}}t(t^2-1)dt=\left ( \frac{2t^5}{5}-\frac{2t^3}{3} \right ) \bigg |_{\sqrt{2}}^{\sqrt{3}}=\frac{8}{5}\sqrt{3}-\frac{4}{15}\sqrt{2}\) Tính \(\int_{1}^{2}xlnxdx=\frac{x^2}{2}lnx \bigg|_{1}^{2}-\int_{1}^{2}\frac{x}{2}dx=2ln2-\frac{x^2}{4} \bigg|_{1}^{2}= 2ln2-\frac{3}{4}\) Vậy \(I=I_1+I_2=\frac{8}{5}\sqrt{5}-\frac{4}{15}\sqrt{2}+2ln2-\frac{3}{4}\)
