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Câu hỏi: Tích phân $I=\int_0^2 \dfrac{x+4}{x^2+3 x+2} \mathrm{~d} x=a \ln 3+b \ln 2$. Khi đó $b^2-a$ bằng
A. $b^2-a=-4$.
B. $b^2-a=-1$.
C. $b^2-a=0$.
D. $b^2-a=1$.
A. $b^2-a=-4$.
B. $b^2-a=-1$.
C. $b^2-a=0$.
D. $b^2-a=1$.
Ta có: $\int_0^2 \dfrac{x+4}{x^2+3 x+2} \mathrm{~d} x=\int_0^2 \dfrac{x+4}{(x+1)(x+2)} \mathrm{d} x=\int_0^2\left(\dfrac{3}{x+1}-\dfrac{2}{x+2}\right) \mathrm{d} x$
$
\begin{aligned}
& =\left.(3 \ln |x+1|-2 \ln |x+2|)\right|_0 ^2=3 \ln 3-2 \ln 4-(-2 \ln 2)=3 \ln 3-2 \ln 2 \\
& \Rightarrow\left\{\begin{array}{l}
a=3 \\
b=-2
\end{array}\right.
\end{aligned}
$
Do đó: $b^2-a=1$.
$
\begin{aligned}
& =\left.(3 \ln |x+1|-2 \ln |x+2|)\right|_0 ^2=3 \ln 3-2 \ln 4-(-2 \ln 2)=3 \ln 3-2 \ln 2 \\
& \Rightarrow\left\{\begin{array}{l}
a=3 \\
b=-2
\end{array}\right.
\end{aligned}
$
Do đó: $b^2-a=1$.
Đáp án D.