$\displaystyle \int_{2}^{8} \frac{1}{x(log_8x)^2}\,dx$
If you let $\displaystyle t=log_8x$, then $\displaystyle dt=\frac{1}{xln(8)}$.
Thus you have $\displaystyle \int_{2}^{8}\frac{ln(8)\,dt}{t^2}$.
Pull out the $\displaystyle ln(8)$ to get $\displaystyle ln(8)\int_{2}^{8}\frac{\,dt}{t^2} = ln(8)*-\frac{1}{t} = ln(8)*-\frac{1}{log_8x} = -\frac{ln(8)}{log_8x}$.
Plugging in $\displaystyle 8$ gives you $\displaystyle -\frac{ln(8)}{log_88} = -\frac{ln(8)}{1}=-ln(8)$ and plugging in $\displaystyle 2$ gives you $\displaystyle -\frac{ln(8)}{log_82} = -\frac{ln(8)}{\frac{1}{3}} = ln(8)*-3 = -3ln(8)$.
Subtracting gives: $\displaystyle -ln(8)+3ln(8) = 2ln(8) = 2ln(2^3) = 6ln(2)$.
Thus, $\displaystyle \int_{2}^{8} \frac{1}{x(log_8x)^2}\,dx = 6ln(2)$.