Cho hàm số y = f(x) liên tục và có đạo hàm trên $\left(0;+\infty \right)$, có đồ thị như hình vẽ đồng thời thỏa mãn $f^{\text{'}}\left(x\right)-\frac{1}{x^2}{f}^{\text{'}}\left(\frac{1}{x}\right)=\frac{5}{18}\left(1-\frac{1}{x^2}\right),\forall x>0$. Diện tích hình phẳng giới hạn bởi các đường $y=\frac{f\left(x\right)-{\left(x-1\right)}^2}{x}$ và y = 0 bằng

Đáp án đúng là: C

Xét phương trình $y=\frac{f\left(x\right)-{\left(x-1\right)}^2}{x}=0\iff f\left(x\right)={\left(x-1\right)}^2\iff \left[\begin{array}{l}x=\frac{1}{2}\\ x=2.\end{array}\right.$

Khi đó $S=\left|\underset{\frac{1}{2}}{\overset{2}{\int }}\frac{f\left(x\right)-{\left(x-1\right)}^2}{x}\text{d}x\right|=\left|\underset{\frac{1}{2}}{\overset{2}{\int }}\frac{f\left(x\right)}{x}\text{d}x-\underset{\frac{1}{2}}{\overset{2}{\int }}\left(x-2+\frac{1}{x}\right)\text{d}x\right|=\left|A-B\right|$.

Tính $B=\underset{\frac{1}{2}}{\overset{2}{\int }}\left(x-2+\frac{1}{x}\right)\text{d}x={\left.\left(\frac{x^2}{2}-2x+\ln x\right)\right|}_{\frac{1}{2}}^2=-\frac{9}{8}+2\ln 2$.

Tính $A=\underset{\frac{1}{2}}{\overset{2}{\int }}\frac{f\left(x\right)}{x}\text{d}x=\underset{\frac{1}{2}}{\overset{2}{\int }}f\left(x\right)\text{d}\left(\ln x\right)={\left.f\left(x\right)\ln x\right|}_{\frac{1}{2}}^2-\underset{\frac{1}{2}}{\overset{2}{\int }}{f}^{\text{'}}\left(x\right)\ln x\text{d}x=\frac{5}{4}\ln 2-\underset{\frac{1}{2}}{\overset{2}{\int }}{f}^{\text{'}}\left(x\right)\ln x\text{d}x$.

Xét phương trình

$f^{\text{'}}\left(x\right)-\frac{1}{x^2}{f}^{\text{'}}\left(\frac{1}{x}\right)=\frac{5}{18}\left(1-\frac{1}{x^2}\right),\forall x>0\iff f^{\text{'}}\left(x\right)\ln x-\frac{1}{x^2}{f}^{\text{'}}\left(\frac{1}{x}\right)\ln x=\frac{5}{18}\left(1-\frac{1}{x^2}\right)\ln x$.

Suy ra $\underset{\frac{1}{2}}{\overset{2}{\int }}{f}^{\text{'}}\left(x\right)\ln x\text{d}x+\underset{\frac{1}{2}}{\overset{2}{\int }}-\frac{1}{x^2}{f}^{\text{'}}\left(\frac{1}{x}\right)\ln x\text{d}x=\frac{5}{18}\underset{\frac{1}{2}}{\overset{2}{\int }}\left(1-\frac{1}{x^2}\right)\ln x\text{d}x$.

Đặt $t=\frac{1}{x}\Rightarrow \text{d}t=-\frac{1}{x^2}\text{d}x$, ta có $x=\frac{1}{2}\Rightarrow t=2$, $x=2\Rightarrow t=\frac{1}{2}$.

Khi đó $\underset{\frac{1}{2}}{\overset{2}{\int }}-\frac{1}{x^2}{f}^{\text{'}}\left(\frac{1}{x}\right)\ln x\text{d}x=\underset{2}{\overset{\frac{1}{2}}{\int }}{f}^{\text{'}}\left(t\right)\ln \frac{1}{t}\text{d}t=\underset{\frac{1}{2}}{\overset{2}{\int }}{f}^{\text{'}}\left(t\right)\ln t\text{d}t=\underset{\frac{1}{2}}{\overset{2}{\int }}{f}^{\text{'}}\left(x\right)\ln x\text{d}x$.

Lại có

$\underset{\frac{1}{2}}{\overset{2}{\int }}\left(1-\frac{1}{x^2}\right)\ln x\text{d}x=\underset{\frac{1}{2}}{\overset{2}{\int }}\ln x\text{d}\left(x+\frac{1}{x}\right)={\left.\left(x+\frac{1}{x}\right)\ln x\right|}_{\frac{1}{2}}^2-\underset{\frac{1}{2}}{\overset{2}{\int }}\left(x+\frac{1}{x}\right)\frac{1}{x}\text{d}x=5\ln 2-3$.

Suy ra $2\underset{\frac{1}{2}}{\overset{2}{\int }}{f}^{\text{'}}\left(x\right)\ln x\text{d}x=\frac{5}{18}\left(5\ln 2-3\right)\iff \underset{\frac{1}{2}}{\overset{2}{\int }}{f}^{\text{'}}\left(x\right)\ln x\text{d}x=\frac{25}{36}\ln 2-\frac{5}{12}$.

Do đó $A=\frac{5}{4}\ln 2-\left(\frac{25}{36}\ln 2-\frac{5}{12}\right)=\frac{5}{9}\ln 2+\frac{5}{12}$.

Vậy $S=\left|A-B\right|=\left|\frac{5}{9}\ln 2+\frac{5}{12}+\frac{9}{8}-2\ln 2\right|=\frac{37}{24}-\frac{13}{9}\ln 2$.