$64k$

$\begin{aligned} \int_1^5 \frac {dx}{2x-1} & = \frac 12 \ln (2x-1) \bigg|_1^5 \\ & = \frac 12 (\ln (2(5)-1) - \ln (2(1)-1)) \\ & = \frac 12 \left(\ln 9 - \ln 1\right) \\ & = \frac 12 \left(\ln 9 - 0 \right) \\ & = \ln 9^\frac 12 \\ & = \ln \sqrt 9 \\ & = \ln 3 \end{aligned}$

$\implies k = \boxed{3}$

$\large \displaystyle\int^{5}_{1} \dfrac{dx}{2x-1}=\ln k$

Let us substitute $2x-1 = u \Rightarrow \,dx = \frac{1}{2}\,du$ $\displaystyle\int^{5}_{1}\frac{1}{2}\dfrac{\,du}{u}=\ln k$ $\displaystyle\dfrac{1}{2} \int^{5}_{1} \ln(u)=\ln k$ $\displaystyle\dfrac{1}{2} \int^{5}_{1}\ln(2x-1) = \ln k$ $\frac {1}{2}\ln (2x-1) \bigg|_1^5 = \ln k$ $\dfrac{1}{2}\left[\ln(9)- \ln (1)\right] = \ln k$ $\ln\sqrt{9}=\ln k\Rightarrow k= 3$

Therefore, $64k = 64\times3 = \boxed{192}$

But weird why we need to know 64k, but here we go:

The primitive is $\frac{1}{2} ln(2x - 1)$ , so if we fill in the boundaries we get:

$\frac{1}{2} ln(9) - \frac{1}{2} ln(0) = ln(3)$ . So k =3 and 64k = 192.