evaluated from upper bound 4 to lower bound 1
Next step here?
That is correct so far (don't forget your dy on the integral)...
$\displaystyle \begin{align*} L &= \int_1^4{\sqrt{ 1 + \frac{1}{4}y^{-1} - 8y^{-\frac{1}{2}} + 64}\,\mathrm{d}y } \\ &= \int_1^4{ \sqrt{ 65 - \frac{8}{\sqrt{y}} + \frac{1}{4y} } \, \mathrm{d}y } \\ &= \int_1^4{\sqrt{ \frac{260y - 32\sqrt{y} + 1}{4y}}\,\mathrm{d}y } \\ &= \int_1^4{ \frac{\sqrt{260y - 32\sqrt{y} + 1}}{\sqrt{4y}}\,\mathrm{d}y } \\ &= \int_1^4{ \sqrt{260 \left( \sqrt{y} \right) ^2 - 32\sqrt{y} + 1 } \, \left( \frac{1}{2\sqrt{y}} \right) \, \mathrm{d}y } \end{align*}$
So now substitute $\displaystyle \begin{align*} u = \sqrt{y} \implies \mathrm{d}u = \frac{1}{2\sqrt{y}}\,\mathrm{d}y \end{align*}$ and noting that $\displaystyle \begin{align*} u(1) = 1 \end{align*}$ and $\displaystyle \begin{align*} u(4) = 2 \end{align*}$, the integral becomes
$\displaystyle \begin{align*} \int_1^4{ \sqrt{ 260 \left( \sqrt{y} \right) ^2 - 32\sqrt{y} + 1 } \, \left( \frac{1}{2\sqrt{y}} \right) \, \mathrm{d}y } &= \int_1^2{ \sqrt{ 260u^2 - 32u + 1} \, \mathrm{d}u } \\ &= \int_1^2{ \sqrt{ 260 \left( u^2 - \frac{8}{65} u + \frac{1}{260} \right) } \, \mathrm{d}u } \\ &= \sqrt{260} \int_1^2{ \sqrt{ u^2 - \frac{8}{65}u + \frac{1}{260} } \, \mathrm{d}u } \\ &= 2\sqrt{65} \int_1^2{ \sqrt{ u^2 - \frac{8}{65}u + \left( -\frac{4}{65} \right) ^2 - \left( - \frac{4}{65} \right) ^2 + \frac{1}{260 } }\,\mathrm{d}u } \\ &= 2\sqrt{65} \int_1^2{\sqrt{ \left( u - \frac{4}{65} \right) ^2 + \frac{1}{16900} } \, \mathrm{d}u} \end{align*}$
So now you can proceed by substituting $\displaystyle \begin{align*} u - \frac{4}{65} = \frac{1}{130}\tan{ \left( \theta \right) } \implies \mathrm{d}u = \frac{1}{130}\sec^2{ \left( \theta \right) } \,\mathrm{d}\theta \end{align*}$.