Google
Câu hỏi: Cho $\int\limits_{1}^{4}{\dfrac{1}{2\sqrt{x}}{{\left( \dfrac{\sqrt{x}+2}{\sqrt{x}+1} \right)}^{2}}d\text{x}}=\dfrac{a}{b}+2\ln \dfrac{c}{d}$ với a, b, c, d là các số nguyên, $\dfrac{a}{b}$ và $\dfrac{c}{d}$ là các phân số tối giản. Giá trị của $a+b+c+d$ bằng:
A. 16
B. 18
C. 25
D. 20
A. 16
B. 18
C. 25
D. 20
Đặt $t=\sqrt{x}\Rightarrow dt=\dfrac{d\text{x}}{2\sqrt{x}}$. Đổi cận $\left\{ \begin{aligned}
& x=1\Rightarrow t=1 \\
& x=4\Rightarrow t=2 \\
\end{aligned} \right.$
$\Rightarrow \int\limits_{1}^{4}{\dfrac{1}{2\sqrt{x}}{{\left( \dfrac{\sqrt{x}+2}{\sqrt{x}+1} \right)}^{2}}d\text{x}}=\int\limits_{1}^{2}{{{\left( \dfrac{t+2}{t+1} \right)}^{2}}dt}=\int\limits_{1}^{2}{{{\left( 1+\dfrac{1}{t+1} \right)}^{2}}dt}$
$=\int\limits_{1}^{2}{\left( 1+\dfrac{2}{t+1}+\dfrac{1}{{{\left( t+1 \right)}^{2}}} \right)dt}=\left. \left( t+2\ln \left| t+1 \right|-\dfrac{1}{t+1} \right) \right|_{1}^{2}$
$=2+2\ln 3-\dfrac{1}{3}-1-2\ln 2+\dfrac{1}{2}=\dfrac{7}{6}+2\ln \dfrac{3}{2}=\dfrac{a}{b}+2\ln \dfrac{c}{d}$
$\Rightarrow a=7;b=6;c=3;d=2\Rightarrow a+b+c+d=7+6+3+2=18$.
& x=1\Rightarrow t=1 \\
& x=4\Rightarrow t=2 \\
\end{aligned} \right.$
$\Rightarrow \int\limits_{1}^{4}{\dfrac{1}{2\sqrt{x}}{{\left( \dfrac{\sqrt{x}+2}{\sqrt{x}+1} \right)}^{2}}d\text{x}}=\int\limits_{1}^{2}{{{\left( \dfrac{t+2}{t+1} \right)}^{2}}dt}=\int\limits_{1}^{2}{{{\left( 1+\dfrac{1}{t+1} \right)}^{2}}dt}$
$=\int\limits_{1}^{2}{\left( 1+\dfrac{2}{t+1}+\dfrac{1}{{{\left( t+1 \right)}^{2}}} \right)dt}=\left. \left( t+2\ln \left| t+1 \right|-\dfrac{1}{t+1} \right) \right|_{1}^{2}$
$=2+2\ln 3-\dfrac{1}{3}-1-2\ln 2+\dfrac{1}{2}=\dfrac{7}{6}+2\ln \dfrac{3}{2}=\dfrac{a}{b}+2\ln \dfrac{c}{d}$
$\Rightarrow a=7;b=6;c=3;d=2\Rightarrow a+b+c+d=7+6+3+2=18$.
Đáp án B.