Cho hàm số y = f(x)  liên tục trên $\left[\frac{1}{3};3\right]$  thỏa mãn $f\left(x\right)+x\cdot f\left(\frac{1}{x}\right)=x^3-x$ . Giá trị tích phân $I=\underset{\frac{1}{3}}{\overset{3}{\int }}\frac{f\left(x\right)}{x^2+x}\text{d}x$  bằng

Đáp án đúng là: D.

Ta có $f\left(x\right)+x\cdot f\left(\frac{1}{x}\right)=x^3-x\iff \frac{f\left(x\right)}{x^2+x}+\frac{1}{x+1}\cdot f\left(\frac{1}{x}\right)=x-1$ .

Suy ra $\underset{\frac{1}{3}}{\overset{3}{\int }}\frac{f\left(x\right)}{x^2+x}\text{d}x+\underset{\frac{1}{3}}{\overset{3}{\int }}\frac{1}{x+1}\cdot f\left(\frac{1}{x}\right)\text{d}x=\underset{\frac{1}{3}}{\overset{3}{\int }}\left(x-1\right)\text{d}x=\frac{16}{9}\left(*\right)$

Đặt $t=\frac{1}{x}$  suy ra $\text{d}x=-\frac{1}{t^2}\text{d}t$  và $x=\frac{1}{t}\iff \frac{1}{x+1}=\frac{t}{t+1}$ .

Đổi cận $\left\{\begin{array}{l}x=\frac{1}{3}\Rightarrow t=3\\ x=3\Rightarrow t=\frac{1}{3}\end{array}\right.$ .

 Khi đó $\underset{\frac{1}{3}}{\overset{3}{\int }}\frac{1}{x+1}\cdot f\left(\frac{1}{x}\right)\text{d}x=-\underset{3}{\overset{\frac{1}{3}}{\int }}\frac{t}{t+1}\cdot f\left(t\right)\cdot \frac{1}{t^2}\text{d}t=\underset{\frac{1}{3}}{\overset{3}{\int }}\frac{f\left(t\right)}{t^2+t}\text{d}t=I$ .

Do đó $\left(*\right)\iff 2I=\frac{16}{9}\iff I=\frac{8}{9}$ .