If $a f(x)+b f\left(\frac{1}{x}\right)=\frac{1}{x}-5, x \neq 0, a \neq b$, then $\int\limits_1^2 f(x) d x$ equals

$\frac{(\ln 2-5) a+\frac{7 b}{2}}{a^2-b^2}$

We have,

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

$\Rightarrow b f(x)+a f\left(\frac{1}{x}\right)=x-5$         [Replacing x by $\frac{1}{x}$]

Solving these equations for f(x), we get

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

∴  $\int\limits_1^2 f(x) d x=\frac{1}{a^2-b^2}\left[a \ln x-b \frac{x^2}{2}-5(a-b) x\right]_1^2$

$\Rightarrow \int\limits_1^2 f(x) d x=\frac{1}{a^2-b^2}\left[a \ln 2-\frac{3 b}{2}-5(a-b)\right]$

$\Rightarrow \int\limits_1^2 f(x) d x=\frac{1}{a^2-b^2}\left[(\ln 2-5) a+\frac{7 b}{2}\right]$