Google
Câu hỏi: Cho $\int\limits_{1}^{3}{\dfrac{3+\ln x}{{{\left( x+1 \right)}^{2}}}dx}=a\ln 3+b\ln 2+c$ với a, b, c là các số hữu tỉ. Giá trị của ${{a}^{2}}+{{b}^{2}}-{{c}^{2}}$ bằng
A. $\dfrac{17}{18}$.
B. $\dfrac{1}{8}$.
C. 1.
D. 0.
A. $\dfrac{17}{18}$.
B. $\dfrac{1}{8}$.
C. 1.
D. 0.
$I=\int\limits_{1}^{3}{\dfrac{3+\ln x}{{{\left( x+1 \right)}^{2}}}dx}$
Đặt $\left\{ \begin{aligned}
& u=3+\ln x \\
& dv=\dfrac{dx}{{{\left( x+1 \right)}^{2}}} \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& du=\dfrac{dx}{x} \\
& v=\dfrac{-1}{x+1} \\
\end{aligned} \right.$.
Khi đó ta có: $I=-\dfrac{3+\ln x}{x+1}\left| \begin{aligned}
& 3 \\
& 1 \\
\end{aligned} \right.+\int\limits_{1}^{3}{\dfrac{dx}{x\left( x+1 \right)}=-}\dfrac{3+\ln x}{x+1}\left| \begin{aligned}
& 3 \\
& 1 \\
\end{aligned} \right.+\ln x\left| \begin{aligned}
& 3 \\
& 1 \\
\end{aligned} \right.-\ln \left( x+1 \right)\left| \begin{aligned}
& 3 \\
& 1 \\
\end{aligned} \right.$
$=\dfrac{3}{4}\ln 3-\ln 2+\dfrac{3}{4}$
Suy ra $\left\{ \begin{aligned}
& a=\dfrac{3}{4} \\
& b=-1 \\
& c=\dfrac{3}{4} \\
\end{aligned} \right.\Rightarrow {{a}^{2}}+{{b}^{2}}-{{c}^{2}}=1$
Đặt $\left\{ \begin{aligned}
& u=3+\ln x \\
& dv=\dfrac{dx}{{{\left( x+1 \right)}^{2}}} \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& du=\dfrac{dx}{x} \\
& v=\dfrac{-1}{x+1} \\
\end{aligned} \right.$.
Khi đó ta có: $I=-\dfrac{3+\ln x}{x+1}\left| \begin{aligned}
& 3 \\
& 1 \\
\end{aligned} \right.+\int\limits_{1}^{3}{\dfrac{dx}{x\left( x+1 \right)}=-}\dfrac{3+\ln x}{x+1}\left| \begin{aligned}
& 3 \\
& 1 \\
\end{aligned} \right.+\ln x\left| \begin{aligned}
& 3 \\
& 1 \\
\end{aligned} \right.-\ln \left( x+1 \right)\left| \begin{aligned}
& 3 \\
& 1 \\
\end{aligned} \right.$
$=\dfrac{3}{4}\ln 3-\ln 2+\dfrac{3}{4}$
Suy ra $\left\{ \begin{aligned}
& a=\dfrac{3}{4} \\
& b=-1 \\
& c=\dfrac{3}{4} \\
\end{aligned} \right.\Rightarrow {{a}^{2}}+{{b}^{2}}-{{c}^{2}}=1$
Đáp án C.