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Câu hỏi: Diện tích hình phẳng giới hạn bởi hàm số $y={{x}^{2}}\sqrt{{{x}^{2}}+1}$, trục Ox và đường thẳng $x=1$ bằng $\dfrac{a\sqrt{b}-\ln \left( 1+\sqrt{b} \right)}{c}$ với a, b, c là các số nguyên dương. Khi đó giá trị của $a+b+c$ là
A. 11
B. 12
C. 13
D. 14
Cách 1: Giải tự luận
Phương trình hoành độ giao điểm ${{x}^{2}}\sqrt{{{x}^{2}}+1}=0\Leftrightarrow x=0$
Diện tích hình phẳng cần tìm là $S=\int\limits_{0}^{1}{{{x}^{2}}\sqrt{{{x}^{2}}+1}dx}$ vì ${{x}^{2}}\sqrt{{{x}^{2}}+1}\ge 0, \forall x\in \left[ 0; 1 \right]$
Đặt $x=\tan t\Rightarrow dx=\left( 1+{{\tan }^{2}}t \right)dt$
Đổi cận $x=0\Rightarrow t=0; x=1\Rightarrow t=\dfrac{\pi }{4}$
Khi đó $S=\int\limits_{0}^{\dfrac{\pi }{4}}{{{\tan }^{2}}t\sqrt{1+{{\tan }^{2}}t}\left( 1+{{\tan }^{2}}t \right)dt}=\int\limits_{0}^{\dfrac{\pi }{4}}{\dfrac{{{\sin }^{2}}t}{{{\cos }^{2}}t}\dfrac{1}{\cos t}.\dfrac{1}{{{\cos }^{2}}t}dt}=\int\limits_{0}^{\dfrac{\pi }{4}}{\dfrac{{{\sin }^{2}}t.\cos t}{{{\left( {{\cos }^{2}}t \right)}^{3}}}dt}$
Đặt $u=\sin t\Rightarrow du=\cos tdt$
Đổi cận $t=0\Rightarrow u=0; t=\dfrac{\pi }{4}\Rightarrow u=\dfrac{\sqrt{2}}{2}$
$S=\int\limits_{0}^{\dfrac{\sqrt{2}}{2}}{\dfrac{{{u}^{2}}}{{{\left( 1-{{u}^{2}} \right)}^{3}}}du}=\int\limits_{0}^{\dfrac{\sqrt{2}}{2}}{\dfrac{1-\left( 1-{{u}^{2}} \right)}{{{\left( 1-{{u}^{2}} \right)}^{3}}}du}=\int\limits_{0}^{\dfrac{\sqrt{2}}{2}}{\left( \dfrac{1}{{{\left( 1-{{u}^{2}} \right)}^{3}}}-\left( \dfrac{1}{{{\left( 1-{{u}^{2}} \right)}^{2}}} \right) \right)du}$
Ta có $H=\int\limits_{0}^{\dfrac{\sqrt{2}}{2}}{\dfrac{1}{{{\left( 1-{{u}^{2}} \right)}^{3}}}du}=\dfrac{1}{8}\int\limits_{0}^{\dfrac{\sqrt{2}}{2}}{{{\left( \dfrac{1-u+1+u}{\left( 1-u \right)\left( 1+u \right)} \right)}^{3}}du}=\dfrac{1}{8}\int\limits_{0}^{\dfrac{\sqrt{2}}{2}}{{{\left( \dfrac{1}{1+u}+\dfrac{1}{1-u} \right)}^{3}}du}$
$=\dfrac{1}{8}\int\limits_{0}^{\dfrac{\sqrt{2}}{2}}{\left( \dfrac{1}{{{\left( 1+u \right)}^{3}}}+\dfrac{1}{{{\left( 1-u \right)}^{3}}}+\dfrac{3}{1-{{u}^{2}}}\left( \dfrac{1}{1-u}+\dfrac{1}{1+u} \right) \right)du}$
$=\dfrac{1}{8}\int\limits_{0}^{\dfrac{\sqrt{2}}{2}}{\left( \dfrac{1}{{{\left( 1+u \right)}^{3}}}+\dfrac{1}{{{\left( 1-u \right)}^{3}}}+\dfrac{6}{{{\left( 1-{{u}^{2}} \right)}^{2}}} \right)du}$
$=\left( \dfrac{-1}{16{{\left( 1+u \right)}^{2}}}+\dfrac{1}{16{{\left( 1-u \right)}^{2}}} \right)\left| \begin{aligned}
& ^{\dfrac{\sqrt{2}}{2}} \\
& _{0} \\
\end{aligned} \right.+\dfrac{1}{8}\int\limits_{0}^{\dfrac{\sqrt{2}}{2}}{\dfrac{6}{{{\left( 1-{{u}^{2}} \right)}^{2}}}du}=\dfrac{\sqrt{2}}{2}+\dfrac{1}{8}\int\limits_{0}^{\dfrac{\sqrt{2}}{2}}{\dfrac{6}{{{\left( 1-{{u}^{2}} \right)}^{2}}}}du$
Tính $K=\int\limits_{0}^{\dfrac{\sqrt{2}}{2}}{\dfrac{6}{{{\left( 1-{{u}^{2}} \right)}^{2}}}du}$ :
$K=\int\limits_{0}^{\dfrac{\sqrt{2}}{2}}{\dfrac{6}{{{\left( 1-{{u}^{2}} \right)}^{2}}}du=\dfrac{3}{2}\int\limits_{0}^{\dfrac{\sqrt{2}}{2}}{{{\left( \dfrac{1-u+1+u}{\left( 1-u \right)\left( 1+u \right)} \right)}^{2}}du}}=\dfrac{3}{2}\int\limits_{0}^{\dfrac{\sqrt{2}}{2}}{{{\left( \dfrac{1}{1-u}+\dfrac{1}{1+u} \right)}^{2}}du}$
$=\dfrac{3}{2}\int\limits_{0}^{\dfrac{\sqrt{2}}{2}}{\left( \dfrac{1}{{{\left( 1-u \right)}^{2}}}+\dfrac{1}{{{\left( 1+u \right)}^{2}}}+\dfrac{2}{\left( 1-u \right)\left( 1+u \right)} \right)du}$
$=\dfrac{3}{2}\left( \dfrac{1}{1-u}-\dfrac{1}{1+u}+\ln \left| \dfrac{1+u}{1-u} \right| \right)\left| \begin{aligned}
& ^{\dfrac{\sqrt{2}}{2}} \\
& _{0} \\
\end{aligned} \right.=3\sqrt{2}+3\ln \left( 1+\sqrt{2} \right)$
Vậy $H=\dfrac{\sqrt{2}}{2}+\dfrac{3\sqrt{2}+3\ln \left( 1+\sqrt{2} \right)}{8}=\dfrac{7\sqrt{2}+3\ln \left( 1+\sqrt{2} \right)}{8}$
Khi đó $S=\dfrac{7\sqrt{2}+3\ln \left( 1+\sqrt{2} \right)}{8}-\dfrac{1}{6}K$
$=\dfrac{7\sqrt{2}+3\ln \left( 1+\sqrt{2} \right)}{8}-\dfrac{1}{6}\left( 3\sqrt{2}+3\ln \left( 1+\sqrt{2} \right) \right)=\dfrac{3\sqrt{2}-\ln \left( 1+\sqrt{2} \right)}{8}$
Cách 2: $S=\int\limits_{0}^{1}{{{x}^{2}}\sqrt{{{x}^{2}}+1}dx}$
Đặt $u-x=\sqrt{{{x}^{2}}+1}\Rightarrow {{u}^{2}}-2ux+{{x}^{2}}={{x}^{2}}+1\Rightarrow x=\dfrac{{{u}^{2}}-1}{2u}\Rightarrow dx=\dfrac{1}{2}\left( 1+\dfrac{1}{{{u}^{2}}} \right)du$
Đổi cận: Với $x=0\Rightarrow u=1$ ; với $x=1\Rightarrow u=1+\sqrt{2}$
Suy ra $S=\dfrac{1}{2}\int\limits_{1}^{1+\sqrt{2}}{{{\left( \dfrac{{{u}^{2}}-1}{2u} \right)}^{2}}}\sqrt{{{\left( \dfrac{{{u}^{2}}-1}{2u} \right)}^{2}}+1}.\left( 1+\dfrac{1}{{{u}^{2}}} \right)du$
$=\dfrac{1}{2}{{\int\limits_{1}^{1+\sqrt{2}}{\left( \dfrac{{{u}^{2}}-1}{2u} \right)}}^{2}}\sqrt{{{\left( \dfrac{{{u}^{2}}-1}{2u} \right)}^{2}}+1}.\left( 1+\dfrac{1}{{{u}^{2}}} \right)du$
$=\dfrac{1}{2}\int\limits_{1}^{1+\sqrt{2}}{\dfrac{{{\left( {{u}^{2}}-1 \right)}^{2}}}{4{{u}^{2}}}.\dfrac{\left( {{u}^{2}}+1 \right)}{2u}.\dfrac{{{u}^{2}}+1}{{{u}^{2}}}du}=\dfrac{1}{16}\int\limits_{1}^{1+\sqrt{2}}{\dfrac{{{\left( {{u}^{4}}-1 \right)}^{2}}}{{{u}^{5}}}du}=\dfrac{1}{16}\int\limits_{1}^{1+\sqrt{2}}{\dfrac{{{u}^{8}}-2{{u}^{4}}+1}{{{u}^{5}}}du}$
$=\dfrac{1}{16}\int\limits_{1}^{1+\sqrt{2}}{\left( {{u}^{3}}-\dfrac{2}{u}+\dfrac{1}{{{u}^{5}}} \right)du}$
$=\dfrac{1}{16}\left( \dfrac{{{u}^{4}}}{4}-2\ln \left| u \right|-\dfrac{1}{4{{u}^{4}}} \right)\left| \begin{aligned}
& ^{1+\sqrt{2}} \\
& _{1} \\
\end{aligned} \right.=\dfrac{1}{8}\left[ \sqrt{18}-\ln \left( 1+\sqrt{2} \right) \right]$
A. 11
B. 12
C. 13
D. 14
Phương trình hoành độ giao điểm ${{x}^{2}}\sqrt{{{x}^{2}}+1}=0\Leftrightarrow x=0$
Diện tích hình phẳng cần tìm là $S=\int\limits_{0}^{1}{{{x}^{2}}\sqrt{{{x}^{2}}+1}dx}$ vì ${{x}^{2}}\sqrt{{{x}^{2}}+1}\ge 0, \forall x\in \left[ 0; 1 \right]$
Đặt $x=\tan t\Rightarrow dx=\left( 1+{{\tan }^{2}}t \right)dt$
Đổi cận $x=0\Rightarrow t=0; x=1\Rightarrow t=\dfrac{\pi }{4}$
Khi đó $S=\int\limits_{0}^{\dfrac{\pi }{4}}{{{\tan }^{2}}t\sqrt{1+{{\tan }^{2}}t}\left( 1+{{\tan }^{2}}t \right)dt}=\int\limits_{0}^{\dfrac{\pi }{4}}{\dfrac{{{\sin }^{2}}t}{{{\cos }^{2}}t}\dfrac{1}{\cos t}.\dfrac{1}{{{\cos }^{2}}t}dt}=\int\limits_{0}^{\dfrac{\pi }{4}}{\dfrac{{{\sin }^{2}}t.\cos t}{{{\left( {{\cos }^{2}}t \right)}^{3}}}dt}$
Đặt $u=\sin t\Rightarrow du=\cos tdt$
Đổi cận $t=0\Rightarrow u=0; t=\dfrac{\pi }{4}\Rightarrow u=\dfrac{\sqrt{2}}{2}$
$S=\int\limits_{0}^{\dfrac{\sqrt{2}}{2}}{\dfrac{{{u}^{2}}}{{{\left( 1-{{u}^{2}} \right)}^{3}}}du}=\int\limits_{0}^{\dfrac{\sqrt{2}}{2}}{\dfrac{1-\left( 1-{{u}^{2}} \right)}{{{\left( 1-{{u}^{2}} \right)}^{3}}}du}=\int\limits_{0}^{\dfrac{\sqrt{2}}{2}}{\left( \dfrac{1}{{{\left( 1-{{u}^{2}} \right)}^{3}}}-\left( \dfrac{1}{{{\left( 1-{{u}^{2}} \right)}^{2}}} \right) \right)du}$
Ta có $H=\int\limits_{0}^{\dfrac{\sqrt{2}}{2}}{\dfrac{1}{{{\left( 1-{{u}^{2}} \right)}^{3}}}du}=\dfrac{1}{8}\int\limits_{0}^{\dfrac{\sqrt{2}}{2}}{{{\left( \dfrac{1-u+1+u}{\left( 1-u \right)\left( 1+u \right)} \right)}^{3}}du}=\dfrac{1}{8}\int\limits_{0}^{\dfrac{\sqrt{2}}{2}}{{{\left( \dfrac{1}{1+u}+\dfrac{1}{1-u} \right)}^{3}}du}$
$=\dfrac{1}{8}\int\limits_{0}^{\dfrac{\sqrt{2}}{2}}{\left( \dfrac{1}{{{\left( 1+u \right)}^{3}}}+\dfrac{1}{{{\left( 1-u \right)}^{3}}}+\dfrac{3}{1-{{u}^{2}}}\left( \dfrac{1}{1-u}+\dfrac{1}{1+u} \right) \right)du}$
$=\dfrac{1}{8}\int\limits_{0}^{\dfrac{\sqrt{2}}{2}}{\left( \dfrac{1}{{{\left( 1+u \right)}^{3}}}+\dfrac{1}{{{\left( 1-u \right)}^{3}}}+\dfrac{6}{{{\left( 1-{{u}^{2}} \right)}^{2}}} \right)du}$
$=\left( \dfrac{-1}{16{{\left( 1+u \right)}^{2}}}+\dfrac{1}{16{{\left( 1-u \right)}^{2}}} \right)\left| \begin{aligned}
& ^{\dfrac{\sqrt{2}}{2}} \\
& _{0} \\
\end{aligned} \right.+\dfrac{1}{8}\int\limits_{0}^{\dfrac{\sqrt{2}}{2}}{\dfrac{6}{{{\left( 1-{{u}^{2}} \right)}^{2}}}du}=\dfrac{\sqrt{2}}{2}+\dfrac{1}{8}\int\limits_{0}^{\dfrac{\sqrt{2}}{2}}{\dfrac{6}{{{\left( 1-{{u}^{2}} \right)}^{2}}}}du$
Tính $K=\int\limits_{0}^{\dfrac{\sqrt{2}}{2}}{\dfrac{6}{{{\left( 1-{{u}^{2}} \right)}^{2}}}du}$ :
$K=\int\limits_{0}^{\dfrac{\sqrt{2}}{2}}{\dfrac{6}{{{\left( 1-{{u}^{2}} \right)}^{2}}}du=\dfrac{3}{2}\int\limits_{0}^{\dfrac{\sqrt{2}}{2}}{{{\left( \dfrac{1-u+1+u}{\left( 1-u \right)\left( 1+u \right)} \right)}^{2}}du}}=\dfrac{3}{2}\int\limits_{0}^{\dfrac{\sqrt{2}}{2}}{{{\left( \dfrac{1}{1-u}+\dfrac{1}{1+u} \right)}^{2}}du}$
$=\dfrac{3}{2}\int\limits_{0}^{\dfrac{\sqrt{2}}{2}}{\left( \dfrac{1}{{{\left( 1-u \right)}^{2}}}+\dfrac{1}{{{\left( 1+u \right)}^{2}}}+\dfrac{2}{\left( 1-u \right)\left( 1+u \right)} \right)du}$
$=\dfrac{3}{2}\left( \dfrac{1}{1-u}-\dfrac{1}{1+u}+\ln \left| \dfrac{1+u}{1-u} \right| \right)\left| \begin{aligned}
& ^{\dfrac{\sqrt{2}}{2}} \\
& _{0} \\
\end{aligned} \right.=3\sqrt{2}+3\ln \left( 1+\sqrt{2} \right)$
Vậy $H=\dfrac{\sqrt{2}}{2}+\dfrac{3\sqrt{2}+3\ln \left( 1+\sqrt{2} \right)}{8}=\dfrac{7\sqrt{2}+3\ln \left( 1+\sqrt{2} \right)}{8}$
Khi đó $S=\dfrac{7\sqrt{2}+3\ln \left( 1+\sqrt{2} \right)}{8}-\dfrac{1}{6}K$
$=\dfrac{7\sqrt{2}+3\ln \left( 1+\sqrt{2} \right)}{8}-\dfrac{1}{6}\left( 3\sqrt{2}+3\ln \left( 1+\sqrt{2} \right) \right)=\dfrac{3\sqrt{2}-\ln \left( 1+\sqrt{2} \right)}{8}$
Cách 2: $S=\int\limits_{0}^{1}{{{x}^{2}}\sqrt{{{x}^{2}}+1}dx}$
Đặt $u-x=\sqrt{{{x}^{2}}+1}\Rightarrow {{u}^{2}}-2ux+{{x}^{2}}={{x}^{2}}+1\Rightarrow x=\dfrac{{{u}^{2}}-1}{2u}\Rightarrow dx=\dfrac{1}{2}\left( 1+\dfrac{1}{{{u}^{2}}} \right)du$
Đổi cận: Với $x=0\Rightarrow u=1$ ; với $x=1\Rightarrow u=1+\sqrt{2}$
Suy ra $S=\dfrac{1}{2}\int\limits_{1}^{1+\sqrt{2}}{{{\left( \dfrac{{{u}^{2}}-1}{2u} \right)}^{2}}}\sqrt{{{\left( \dfrac{{{u}^{2}}-1}{2u} \right)}^{2}}+1}.\left( 1+\dfrac{1}{{{u}^{2}}} \right)du$
$=\dfrac{1}{2}{{\int\limits_{1}^{1+\sqrt{2}}{\left( \dfrac{{{u}^{2}}-1}{2u} \right)}}^{2}}\sqrt{{{\left( \dfrac{{{u}^{2}}-1}{2u} \right)}^{2}}+1}.\left( 1+\dfrac{1}{{{u}^{2}}} \right)du$
$=\dfrac{1}{2}\int\limits_{1}^{1+\sqrt{2}}{\dfrac{{{\left( {{u}^{2}}-1 \right)}^{2}}}{4{{u}^{2}}}.\dfrac{\left( {{u}^{2}}+1 \right)}{2u}.\dfrac{{{u}^{2}}+1}{{{u}^{2}}}du}=\dfrac{1}{16}\int\limits_{1}^{1+\sqrt{2}}{\dfrac{{{\left( {{u}^{4}}-1 \right)}^{2}}}{{{u}^{5}}}du}=\dfrac{1}{16}\int\limits_{1}^{1+\sqrt{2}}{\dfrac{{{u}^{8}}-2{{u}^{4}}+1}{{{u}^{5}}}du}$
$=\dfrac{1}{16}\int\limits_{1}^{1+\sqrt{2}}{\left( {{u}^{3}}-\dfrac{2}{u}+\dfrac{1}{{{u}^{5}}} \right)du}$
$=\dfrac{1}{16}\left( \dfrac{{{u}^{4}}}{4}-2\ln \left| u \right|-\dfrac{1}{4{{u}^{4}}} \right)\left| \begin{aligned}
& ^{1+\sqrt{2}} \\
& _{1} \\
\end{aligned} \right.=\dfrac{1}{8}\left[ \sqrt{18}-\ln \left( 1+\sqrt{2} \right) \right]$
Đáp án C.