If $3f\left(x\right)+2f\left(\frac{1}{x}\right)=\frac{1}{x}-5\forall x\ne 0$ , then ${\displaystyle \underset{1}{\overset{2}{\int }}f\left(x\right)dx=}$

$3f\left(x\right)+2f\left(\frac{1}{x}\right)=\frac{1}{x}-5,x\to \frac{1}{x}$

$3f\left(\frac{1}{x}\right)+2f\left(x\right)=x-5$

Eliminating of $f\left(\frac{1}{x}\right)\to 5f\left(x\right)=\frac{3}{x}-15-2x+10=\frac{3}{x}-2x-5$

${\displaystyle \underset{1}{\overset{2}{\int }}f\left(x\right)dx=\frac{1}{5}{\displaystyle \underset{1}{\overset{2}{\int }}\left(\frac{3}{x}-2x-5\right)dx}}$

                    $=\frac{1}{5}\left(3\ln 2-8\right)$