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Câu hỏi: Cho $\int\limits_{0}^{1}{{{\left( \dfrac{2x+1}{x+1} \right)}^{2}}dx=a+b\ln 2}$, với $a,b$ là các số hữu tỉ. Tính $T=4{{a}^{2}}+ab-2{{b}^{2}}$
A. $T=-31.$
B. $T=-28.$
C. $T=31.$
D. $T=28.$
A. $T=-31.$
B. $T=-28.$
C. $T=31.$
D. $T=28.$
Ta có ${{\left( \dfrac{2x+1}{x+1} \right)}^{2}}={{\left( 2-\dfrac{1}{x+1} \right)}^{2}}=4-\dfrac{4}{x+1}+\dfrac{1}{{{\left( x+1 \right)}^{2}}}$
$\Rightarrow \int\limits_{0}^{1}{{{\left( \dfrac{2x+1}{x+1} \right)}^{2}}dx}=\left( 4x-4\ln \left| x+1 \right|-\dfrac{1}{x+1} \right)\left| _{\begin{smallmatrix}
\\
0
\end{smallmatrix}}^{\begin{smallmatrix}
1 \\
\end{smallmatrix}} \right.=\left( 4-4\ln 2-\dfrac{1}{2} \right)+1=\dfrac{9}{2}-4\ln 2$
$\Rightarrow a=\dfrac{9}{2};b=-4\Rightarrow T=31.$
$\Rightarrow \int\limits_{0}^{1}{{{\left( \dfrac{2x+1}{x+1} \right)}^{2}}dx}=\left( 4x-4\ln \left| x+1 \right|-\dfrac{1}{x+1} \right)\left| _{\begin{smallmatrix}
\\
0
\end{smallmatrix}}^{\begin{smallmatrix}
1 \\
\end{smallmatrix}} \right.=\left( 4-4\ln 2-\dfrac{1}{2} \right)+1=\dfrac{9}{2}-4\ln 2$
$\Rightarrow a=\dfrac{9}{2};b=-4\Rightarrow T=31.$
Đáp án C.