Biết $\Large \int\limits_1^2\mathrm{ln}(x+1)\mathrm{d}x=a\mathrm{ln}3+

Biết $\Large \int\limits_1^2\mathrm{ln}(x+1)\mathrm{d}x=a\mathrm{ln}3+b\mathrm{ln}2+c$ với $\Large a, b, c \in \mathbb{Z}.$ Giá trị của biểu thức $\Large S=a+b+c$ là

• A.

  A. $\Large S = 0.$

• B.

  B. $\Large S = -2.$

• C.

  C. $\Large S = 2.$

• D.

  D. $\Large S = 1.$

Chọn A

Đặt: $\Large \left\{\begin{align} & u=\mathrm{ln}(x+1) \\ & \mathrm{d}v=\mathrm{d}x \end{align}\right.$ $\Large \Rightarrow \left\{\begin{align} & \mathrm{d}u=\dfrac{1}{x+1}\mathrm{d}x \\ & v=x+1 \end{align}\right.$

Khi đó: $\Large \int\limits_1^2\mathrm{ln}(x+1)\mathrm{d}x=(x+1)\mathrm{ln}(x+1)\big|_1^2-\int\limits_1^2\mathrm{d}x$ $\Large =3\mathrm{ln}3-2\mathrm{ln}2-x\big|_1^2$ $\Large =3\mathrm{ln}3-2\mathrm{ln}2-1.$

Vậy $\Large S=a+b+c=3+(-2)+(-1)=0.$