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Câu hỏi: Cho hàm số $f\left( x \right)$ có đạo hàm $f'\left( x \right)=\ln \left( x+1 \right),\forall x\in \left( -1;+\infty \right)$. Khi $\int\limits_{0}^{1}{f\left( x \right)}dx=0$ thì $f\left( 0 \right)$ bằng
A. $-\dfrac{5}{4}+2\ln 2$.
B. $\dfrac{3}{4}-2\ln 2$.
C. $\dfrac{5}{4}-2\ln 2$.
D. $-\dfrac{3}{4}+2\ln 2$.
A. $-\dfrac{5}{4}+2\ln 2$.
B. $\dfrac{3}{4}-2\ln 2$.
C. $\dfrac{5}{4}-2\ln 2$.
D. $-\dfrac{3}{4}+2\ln 2$.
Ta có $f'\left( x \right)=\ln \left( x+1 \right),\forall x\in \left( -1;+\infty \right)$
$\begin{aligned}
& \Rightarrow f\left( x \right)=\int{\ln \left( x+1 \right)}dx=x.\ln \left( x+1 \right)-\int{xd\left( \ln (x+1 \right)}=x\ln \left( x+1 \right)-\int{\dfrac{x}{x+1}dx} \\
& =x\ln \left( x+1 \right)-\int{\left( 1-\dfrac{1}{x+1} \right)dx}=x\ln \left( x+1 \right)-x+\ln \left( x+1 \right)+C \\
\end{aligned}$
Khi đó $\int\limits_{0}^{1}{f\left( x \right)}dx=\left. x.f\left( x \right) \right|_{0}^{1}-\int\limits_{0}^{1}{xf'\left( x \right)}dx=f\left( 1 \right)-\int\limits_{0}^{1}{x\ln \left( x+1 \right)}dx=f\left( 1 \right)-\dfrac{1}{2}\left[ \left. {{x}^{2}}\ln \left( x+1 \right) \right|_{0}^{1}-\int\limits_{0}^{1}{\dfrac{{{x}^{2}}}{x+1}}dx \right]$
$=f\left( 1 \right)-\dfrac{1}{2}\ln 2+\left. \dfrac{1}{2}\left( \dfrac{{{x}^{2}}}{2}-x+\ln \left( x+1 \right) \right) \right|_{0}^{1}=f\left( 1 \right)-\dfrac{1}{4}$
$\int\limits_{0}^{1}{f\left( x \right)}dx=0\Rightarrow f\left( 1 \right)=\dfrac{1}{4}=2\ln 2-1+C\Rightarrow C=\dfrac{5}{4}-2\ln 2$
$\Rightarrow f\left( x \right)=x\ln \left( x+1 \right)-x+\ln \left( x+1 \right)+\dfrac{5}{4}-2\ln 2$ $\Rightarrow f\left( 0 \right)=\dfrac{5}{4}-2\ln 2$.
$\begin{aligned}
& \Rightarrow f\left( x \right)=\int{\ln \left( x+1 \right)}dx=x.\ln \left( x+1 \right)-\int{xd\left( \ln (x+1 \right)}=x\ln \left( x+1 \right)-\int{\dfrac{x}{x+1}dx} \\
& =x\ln \left( x+1 \right)-\int{\left( 1-\dfrac{1}{x+1} \right)dx}=x\ln \left( x+1 \right)-x+\ln \left( x+1 \right)+C \\
\end{aligned}$
Khi đó $\int\limits_{0}^{1}{f\left( x \right)}dx=\left. x.f\left( x \right) \right|_{0}^{1}-\int\limits_{0}^{1}{xf'\left( x \right)}dx=f\left( 1 \right)-\int\limits_{0}^{1}{x\ln \left( x+1 \right)}dx=f\left( 1 \right)-\dfrac{1}{2}\left[ \left. {{x}^{2}}\ln \left( x+1 \right) \right|_{0}^{1}-\int\limits_{0}^{1}{\dfrac{{{x}^{2}}}{x+1}}dx \right]$
$=f\left( 1 \right)-\dfrac{1}{2}\ln 2+\left. \dfrac{1}{2}\left( \dfrac{{{x}^{2}}}{2}-x+\ln \left( x+1 \right) \right) \right|_{0}^{1}=f\left( 1 \right)-\dfrac{1}{4}$
$\int\limits_{0}^{1}{f\left( x \right)}dx=0\Rightarrow f\left( 1 \right)=\dfrac{1}{4}=2\ln 2-1+C\Rightarrow C=\dfrac{5}{4}-2\ln 2$
$\Rightarrow f\left( x \right)=x\ln \left( x+1 \right)-x+\ln \left( x+1 \right)+\dfrac{5}{4}-2\ln 2$ $\Rightarrow f\left( 0 \right)=\dfrac{5}{4}-2\ln 2$.
Đáp án C.