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Câu hỏi: Cho hàm số $y=f\left( x \right)$ có đạo hàm liên tục trên đoạn $\left[ 0;\text{1} \right]$ và thỏa mãn $f\left( x \right)+{f}'\left( x \right)=\dfrac{2{{x}^{3}}-5{{x}^{2}}+5x}{{{\left( {{x}^{2}}-x+1 \right)}^{2}}}$ ; $f\left( 1 \right)-f\left( 0 \right)=2$ ; $\int\limits_{0}^{1}{f\left( x \right)\text{d}x=0}$. Biết diện tích hình phẳng giới hạn bởi đồ thị $\left( C \right):y=f\left( x \right)$, trục tung và trục hoành có dạng $S=\ln a-\ln b$ với $a,b$ là các số nguyên dương. Tính $T={{a}^{2}}+{{b}^{2}}$.
A. $T=13$.
B. $T=25$.
C. $T=34$.
D. $T=41$.
A. $T=13$.
B. $T=25$.
C. $T=34$.
D. $T=41$.
Ta có $f\left( x \right)+{f}'\left( x \right)=\dfrac{2{{x}^{3}}-5{{x}^{2}}+5x}{{{\left( {{x}^{2}}-x+1 \right)}^{2}}}$ $=\dfrac{\left( 2x-1 \right)\left( {{x}^{2}}-x+1 \right)-2{{x}^{2}}+2x+1}{{{\left( {{x}^{2}}-x+1 \right)}^{2}}}$
$\Rightarrow \int{f\left( x \right)\text{d}x}+\int{{f}'\left( x \right)\text{d}x}=\int{\dfrac{2x-1}{{{x}^{2}}-x+1}\text{d}x-}\int{\dfrac{2{{x}^{2}}-2x-1}{{{\left( {{x}^{2}}-x+1 \right)}^{2}}}\text{d}x}$
$\Rightarrow \int{f\left( x \right)\text{d}x}+\int{{f}'\left( x \right)\text{d}x}=\int{\dfrac{\text{d}\left( {{x}^{2}}-x+1 \right)}{{{x}^{2}}-x+1}-}\int{\dfrac{\dfrac{2{{x}^{2}}-2x-1}{{{\left( 2x-1 \right)}^{2}}}}{{{\left( \dfrac{{{x}^{2}}-x+1}{2x-1} \right)}^{2}}}\text{d}x}$
$\Rightarrow \int{f\left( x \right)\text{d}x}+f\left( x \right)=\int{\dfrac{\text{d}\left( {{x}^{2}}-x+1 \right)}{{{x}^{2}}-x+1}-}\int{\dfrac{\text{d}\left( \dfrac{{{x}^{2}}-x+1}{2x-1} \right)}{{{\left( \dfrac{{{x}^{2}}-x+1}{2x-1} \right)}^{2}}}}=\ln \left( {{x}^{2}}-x+1 \right)+\dfrac{2x-1}{{{x}^{2}}-x+1}+C$.
Mặt khác, ta có
$\left\{ \begin{aligned}
& \int\limits_{0}^{1}{\dfrac{2x-1}{{{x}^{2}}-x+1}\text{d}x}=\ln \left( {{x}^{2}}-x+1 \right)\left| \begin{aligned}
& 1 \\
& 0 \\
\end{aligned} \right.=0=\int_{0}^{1}{f\left( x \right)\text{d}x} \\
& \left( \dfrac{2x-1}{{{x}^{2}}-x+1} \right)\left| \begin{aligned}
& 1 \\
& 0 \\
\end{aligned} \right.=1-\left( -1 \right)=2=f\left( 1 \right)-f\left( 0 \right) \\
& \int{\dfrac{2x-1}{{{x}^{2}}-x+1}\text{d}x=\ln \left( {{x}^{2}}-x+1 \right)+C} \\
\end{aligned} \right. $ nên suy ra $ \left\{ \begin{aligned}
& C=0 \\
& f\left( x \right)=\dfrac{2x-1}{{{x}^{2}}-x+1} \\
\end{aligned} \right.$.
Do đó $S=\int\limits_{0}^{\dfrac{1}{2}}{\left| \dfrac{2x-1}{{{x}^{2}}-x+1} \right|\text{d}x}=-\ln \left( {{x}^{2}}-x+1 \right)\left| \begin{aligned}
& \dfrac{1}{2} \\
& 0 \\
\end{aligned} \right.=\ln \dfrac{4}{3}=\ln 4-\ln 3 $. Suy ra $ \left\{ \begin{aligned}
& a=4 \\
& b=3 \\
\end{aligned} \right.$.
Vậy $T={{a}^{2}}+{{b}^{2}}=25$.
$\Rightarrow \int{f\left( x \right)\text{d}x}+\int{{f}'\left( x \right)\text{d}x}=\int{\dfrac{2x-1}{{{x}^{2}}-x+1}\text{d}x-}\int{\dfrac{2{{x}^{2}}-2x-1}{{{\left( {{x}^{2}}-x+1 \right)}^{2}}}\text{d}x}$
$\Rightarrow \int{f\left( x \right)\text{d}x}+\int{{f}'\left( x \right)\text{d}x}=\int{\dfrac{\text{d}\left( {{x}^{2}}-x+1 \right)}{{{x}^{2}}-x+1}-}\int{\dfrac{\dfrac{2{{x}^{2}}-2x-1}{{{\left( 2x-1 \right)}^{2}}}}{{{\left( \dfrac{{{x}^{2}}-x+1}{2x-1} \right)}^{2}}}\text{d}x}$
$\Rightarrow \int{f\left( x \right)\text{d}x}+f\left( x \right)=\int{\dfrac{\text{d}\left( {{x}^{2}}-x+1 \right)}{{{x}^{2}}-x+1}-}\int{\dfrac{\text{d}\left( \dfrac{{{x}^{2}}-x+1}{2x-1} \right)}{{{\left( \dfrac{{{x}^{2}}-x+1}{2x-1} \right)}^{2}}}}=\ln \left( {{x}^{2}}-x+1 \right)+\dfrac{2x-1}{{{x}^{2}}-x+1}+C$.
Mặt khác, ta có
$\left\{ \begin{aligned}
& \int\limits_{0}^{1}{\dfrac{2x-1}{{{x}^{2}}-x+1}\text{d}x}=\ln \left( {{x}^{2}}-x+1 \right)\left| \begin{aligned}
& 1 \\
& 0 \\
\end{aligned} \right.=0=\int_{0}^{1}{f\left( x \right)\text{d}x} \\
& \left( \dfrac{2x-1}{{{x}^{2}}-x+1} \right)\left| \begin{aligned}
& 1 \\
& 0 \\
\end{aligned} \right.=1-\left( -1 \right)=2=f\left( 1 \right)-f\left( 0 \right) \\
& \int{\dfrac{2x-1}{{{x}^{2}}-x+1}\text{d}x=\ln \left( {{x}^{2}}-x+1 \right)+C} \\
\end{aligned} \right. $ nên suy ra $ \left\{ \begin{aligned}
& C=0 \\
& f\left( x \right)=\dfrac{2x-1}{{{x}^{2}}-x+1} \\
\end{aligned} \right.$.
Do đó $S=\int\limits_{0}^{\dfrac{1}{2}}{\left| \dfrac{2x-1}{{{x}^{2}}-x+1} \right|\text{d}x}=-\ln \left( {{x}^{2}}-x+1 \right)\left| \begin{aligned}
& \dfrac{1}{2} \\
& 0 \\
\end{aligned} \right.=\ln \dfrac{4}{3}=\ln 4-\ln 3 $. Suy ra $ \left\{ \begin{aligned}
& a=4 \\
& b=3 \\
\end{aligned} \right.$.
Vậy $T={{a}^{2}}+{{b}^{2}}=25$.
Đáp án B.