Tính tích phân sau I = x ln(x+1) dx

Câu hỏi:

Tính tích phân sau $I={\int }_0^{e-1}x\ln \left(x+1\right)dx$

A. $\frac{2-e^2}{6}$

B. $\frac{1+e^3}{4}$

C. $\frac{2-e^2}{4}$

D. $\frac{e^2-3}{4}$

Trả lời:

Chọn D

Đặt $\left\{\begin{array}{l}u=\ln (x+1)\\ dv=xdx\end{array}\Rightarrow \left\{\begin{array}{l}du=\frac{1}{x+1}dx\\ v=\frac{x^2-1}{2}\end{array}\right.\right.$

$$\begin{gathered} I={\int }_0^{e-1}x\ln (x+1)dx \\ =\left[\ln (x+1)\frac{x^2-1}{2}\right]{|}_0^{e-1}-\frac{1}{2}{\int }_0^{e-1}(x-1)dx \\ =\frac{e^2-2e}{2}-\frac{1}{2}\left(\frac{x^2}{2}-x\right){|}_0^{e-1} \\ =\frac{e^2-2e}{2}-\frac{1}{2}\frac{e^2-4e+3}{2}=\frac{e^2-3}{4} \end{gathered}$$