Evaluate the following definite integrals. int_{1/8}^1frac{dx}{xsqrt{1+x^{2/3}}}

$\text{Consider the integral:}$${\int }_{1/8}^1\frac{dx}{x\sqrt{1+x^{2/3}}}$
$\text{Apply u-substitution: }u=\sqrt{1+x^{2/3}}$
${\int }_{1/8}^1\frac{dx}{x\sqrt{1+x^{2/3}}}={\int }_{\frac{5^{1/2}}{2}}^{\sqrt{2}}\frac{3}{u^2-1}du$
$=3\cdot {\int }_{\frac{5^{1/2}}{2}}^{\sqrt{2}}\frac{1}{u^2-1}du$
$=3\cdot {\int }_{\frac{5^{1/2}}{2}}^{\sqrt{2}}\frac{1}{-(-u^2+1)}du$
$=3-{\int }_{\frac{5^{1/2}}{2}}^{\sqrt{2}}\frac{1}{-u^2+1}du$
$=3(-[\frac{\ln |u+1|}{2}-\frac{\ln |u-1|}{2}{]}_{\frac{5^{1/2}}{2}}^{\sqrt{2}})$
$=-3[\frac{1}{2}(\ln |u+1|-\ln |u-1|){]}_{\frac{5^{1/2}}{2}}^{\sqrt{2}}$
$$\begin{gathered} =-3\cdot \frac{\ln (\sqrt{2}+1)-\ln (\sqrt{2}-1)-\ln (\frac{\sqrt{5}}{2}+1)+\ln (\frac{\sqrt{5}}{2}-1)}{2} \\ \text{Answer in terms of logarithms:} \end{gathered}$$
${\int }_{1/8}^1\frac{dx}{x\sqrt{1+x^{2/3}}}=-3\cdot \frac{\ln (\sqrt{2}+1)-\ln (\sqrt{2}-1)-\ln (\frac{\sqrt{5}}{2}+1)+\ln (\frac{\sqrt{5}}{2}-1)}{2}$