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

Biết rằng $\Large \int\limits_1^2 \mathrm{ln}(x+1)\mathrm{d}x=a\mathrm{ln}3+b\mathrm{ln}2+c$ với $\Large a, b, c$ là các số nguyên. Tính $\Large S=a+b+c$ được

• A.

  A. $\Large S=-2.$

• B.

  B. $\Large S=2.$

• C.

  C. $\Large S=1.$

• D.

  D. $\Large S=0.$

Chọn D

$\Large \int\limits_1^2\mathrm{ln}(x+1)\mathrm{d}x$ $\Large =\int\limits_1^2\mathrm{ln}(x+1)\mathrm{d}(x+1)$ $\Large =\big[(x+1)\mathrm{ln}(x+1)\big]\Big|_1^2-\int\limits_1^2(x+1)\mathrm{d}\Big(\mathrm{ln}\big(x+1\big)\Big)$ $\Large=3\mathrm{ln}3-2\mathrm{ln}2-\int\limits_1^2(x+1)\dfrac{1}{x+1}\mathrm{d}x$ $\Large =3\mathrm{ln}3-2\mathrm{ln}2-x\Big|_1^2=3\mathrm{ln}3-2\mathrm{ln}2-1.$

Do đó $\Large a=3, b=-2, c=-1$ $\Large \Rightarrow S=a+b+c=0.$