Explanation

(b)
Given, $2f(x) - 3f\left( \frac{1}{x} \right) = x$ …(i)
Put $x = \frac{1}{x}$ in Eq.(i), we get
$2f\left( \frac{1}{x} \right) - 3f(x) = \frac{1}{x}$ …(ii)
On solving Eqs. (i) and (ii), we get
$f(x) = - \frac{3 + 2x^{2}}{5x}$
Now, $\int_{1}^{2}{f(x)dx} = - \int_{1}^{2}{\frac{3 + 2x^{2}}{5x}dx}$
$= - \int_{1}^{2}{\left( \frac{3}{5x} + \frac{2x}{5} \right)dx}$
$= - \frac{1}{5}\left\lbrack 3\log x + \frac{2x^{2}}{2} \right\rbrack_{1}^{2}$
$= - \frac{1}{5}(3\log 2 + 4 - 3\log{1 - 1)}$
$= - \frac{1}{5}(3\log 2 + 4 - 0 - 1)$
$= - \frac{3}{5}\left( 1 + \log 2 \right)$