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Câu hỏi: Cho hàm số $f\left( x \right)=x+\sqrt{{{x}^{2}}+1}$ biết $\int\limits_{0}^{1}{\dfrac{f\left( x \right)}{f\left( -x \right)}}\text{d}x=a+b\sqrt{c}$ với $a, b, c$ là các số hữu tỷ tối giãn. Tính giá trị $P=a+b+c$.
A. $P=\dfrac{13}{3}$.
B. $P=\dfrac{15}{3}$.
C. $P=\dfrac{10}{3}$.
D. $P=\dfrac{11}{3}$.
A. $P=\dfrac{13}{3}$.
B. $P=\dfrac{15}{3}$.
C. $P=\dfrac{10}{3}$.
D. $P=\dfrac{11}{3}$.
Tập xác định : $D=\mathbb{R}$.
Ta có: $f\left( x \right)=x+\sqrt{{{x}^{2}}+1}\Leftrightarrow f\left( -x \right)=-x+\sqrt{{{x}^{2}}+1}=\dfrac{1}{x+\sqrt{{{x}^{2}}+1}}=\dfrac{1}{f\left( x \right)}$.
Vậy $\dfrac{f\left( x \right)}{f\left( -x \right)}={{\left( x+\sqrt{{{x}^{2}}+1} \right)}^{2}}=2{{x}^{2}}+1+2x\sqrt{{{x}^{2}}+1}$.
Khi đó : $\int\limits_{0}^{1}{\left( 2{{x}^{2}}+1+2x\sqrt{{{x}^{2}}+1} \right)}\text{d}x=\int\limits_{0}^{1}{\left( 2{{x}^{2}}+1 \right)}\text{d}x+\int\limits_{0}^{1}{\left( 2x\sqrt{{{x}^{2}}+1} \right)}\text{d}x=\dfrac{5}{3}+\int\limits_{0}^{1}{\left( {{\left( {{x}^{2}}+1 \right)}^{\prime }}\sqrt{{{x}^{2}}+1} \right)}\text{d}x$
$$ $=\frac{5}{3}+\int\limits_{0}^{1}{\left( \sqrt{{{x}^{2}}+1} \right)}\text{d}\left( {{x}^{2}}+1 \right)=\frac{5}{3}+\left. \frac{2}{3}{{\left( {{x}^{2}}+1 \right)}^{\frac{3}{2}}} \right|_{0}^{1}=\frac{5}{3}+\frac{4\sqrt{2}}{3}-\frac{2}{3}=1+\frac{4}{3}.\sqrt{2}$.
Vậy $a=1 ; b=\frac{4}{3} ; c=2$ khi đó $P=a+b+c=\frac{13}{3}$.
Ta có: $f\left( x \right)=x+\sqrt{{{x}^{2}}+1}\Leftrightarrow f\left( -x \right)=-x+\sqrt{{{x}^{2}}+1}=\dfrac{1}{x+\sqrt{{{x}^{2}}+1}}=\dfrac{1}{f\left( x \right)}$.
Vậy $\dfrac{f\left( x \right)}{f\left( -x \right)}={{\left( x+\sqrt{{{x}^{2}}+1} \right)}^{2}}=2{{x}^{2}}+1+2x\sqrt{{{x}^{2}}+1}$.
Khi đó : $\int\limits_{0}^{1}{\left( 2{{x}^{2}}+1+2x\sqrt{{{x}^{2}}+1} \right)}\text{d}x=\int\limits_{0}^{1}{\left( 2{{x}^{2}}+1 \right)}\text{d}x+\int\limits_{0}^{1}{\left( 2x\sqrt{{{x}^{2}}+1} \right)}\text{d}x=\dfrac{5}{3}+\int\limits_{0}^{1}{\left( {{\left( {{x}^{2}}+1 \right)}^{\prime }}\sqrt{{{x}^{2}}+1} \right)}\text{d}x$
$$ $=\frac{5}{3}+\int\limits_{0}^{1}{\left( \sqrt{{{x}^{2}}+1} \right)}\text{d}\left( {{x}^{2}}+1 \right)=\frac{5}{3}+\left. \frac{2}{3}{{\left( {{x}^{2}}+1 \right)}^{\frac{3}{2}}} \right|_{0}^{1}=\frac{5}{3}+\frac{4\sqrt{2}}{3}-\frac{2}{3}=1+\frac{4}{3}.\sqrt{2}$.
Vậy $a=1 ; b=\frac{4}{3} ; c=2$ khi đó $P=a+b+c=\frac{13}{3}$.
Đáp án A.