$af\left ( x \right )+bf\left ( \frac{1}{x} \right )=\frac{1}{x} -25$ --- $\left ( 1 \right )$
Integrating both sides,
$a\int_{1}^{2}f\left ( x \right )dx+b\int_{1}^{2}f\left ( \frac{1}{x} \right )dx=\left [ \log\left ( x \right )-25x \right ]_{1}^{2}=\log2-25$ --- $\left ( 2 \right )$
Replacing $x$ by $\frac{1}{x}$ in $\left ( 1 \right )$, we get
$af\left ( \frac{1}{x} \right )+bf\left ( x \right )=x-25$
Integrating both sides, we get
$a\int_{1}^{2}f\left ( \frac{1}{x} \right )dx+b\int_{1}^{2}f\left ( x \right )dx=\left [ \frac{x^{2}}{2}-25x \right ]_{1}^{2}=-\frac{47}{2}$ --- $\left ( 3 \right )$
Eliminate $\int_{1}^{2}f\left ( \frac{1}{x} \right )$ between $\left ( 2 \right )$ and $\left ( 3 \right )$ by multiplying $\left ( 2 \right )$ by $a$ and $\left ( 3 \right )$ by $b$ and subtracting
$\therefore \left ( a^{2}-b^{2} \right )\int_{1}^{2}f\left ( x \right )dx=a\left ( \log2-25 \right )+b\times\frac{47}{2}$
$\therefore \int_{1}^{2}f\left ( x \right )dx=\frac{1}{\left ( a^{2}-b^{2} \right )}\left [ a\left ( \log2-25 \right )+\frac{47b}{2} \right ]$
Answer: A. $\frac{1}{\left ( a^{2}-b^{2} \right )}\left [ a\left ( \log2-25 \right )+\frac{47b}{2} \right ]$
