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Câu hỏi: Biết $\int_0^2 2 x \ln (x+1) \mathrm{d} x=a \cdot \ln b$, với $a, b \in \mathbb{N}^*, b$ là số nguyên tố. Tính $6 a+7 b$.
A. 25 .
B. 42 .
C. 39 .
D. 33 .
A. 25 .
B. 42 .
C. 39 .
D. 33 .
Xét $I=\int_0^2 2 x \ln (x+1) \mathrm{d} x=6$.
Đặt $\left\{\begin{array}{l}u=\ln (x+1) \\ \mathrm{d} v=2 x \mathrm{~d} x\end{array} \Leftrightarrow\left\{\begin{array}{l}\mathrm{d} u=\dfrac{1}{x+1} \mathrm{~d} x \\ v=x^2-1\end{array}\right.\right.$.
Ta có $I=\left.\left(x^2-1\right) \ln (x+1)\right|_0 ^2-\int_0^2 \dfrac{x^2-1}{x+1} \mathrm{~d} x=3 \ln 3-\int_0^2(x-1) \mathrm{d} x=3 \ln 3-\left.\left(\dfrac{x^2}{2}-x\right)\right|_0 ^2=3 \ln 3$.
Vậy $a=3, b=3 \Rightarrow 6 a+7 b=39$.
Đặt $\left\{\begin{array}{l}u=\ln (x+1) \\ \mathrm{d} v=2 x \mathrm{~d} x\end{array} \Leftrightarrow\left\{\begin{array}{l}\mathrm{d} u=\dfrac{1}{x+1} \mathrm{~d} x \\ v=x^2-1\end{array}\right.\right.$.
Ta có $I=\left.\left(x^2-1\right) \ln (x+1)\right|_0 ^2-\int_0^2 \dfrac{x^2-1}{x+1} \mathrm{~d} x=3 \ln 3-\int_0^2(x-1) \mathrm{d} x=3 \ln 3-\left.\left(\dfrac{x^2}{2}-x\right)\right|_0 ^2=3 \ln 3$.
Vậy $a=3, b=3 \Rightarrow 6 a+7 b=39$.
Đáp án C.