Answer
$a)$ $6+\ln{2}$
$b)$ $8+\ln{9}$
Work Step by Step
$a)$
$L$ = $\int_{{\,a}}^{{\,b}}\sqrt{1+(\frac{dy}{dx})^{2}}$ $dx$
$\frac{dy}{dx}$ = $\frac{x}{4}-\frac{1}{x}$
$L$ = $\int_{{\,4}}^{{\,8}}\sqrt{1+(\frac{x}{4}-\frac{1}{x})^{2}}$ $dx$
$L$ = $\int_{{\,4}}^{{\,8}}\sqrt{1+\frac{x}{16}^{2}-\frac{1}{2}+\frac{1}{x^{2}}}$ $dx$
$L$ = $\int_{{\,4}}^{{\,8}}\sqrt{\frac{x}{16}^{2}+\frac{1}{2}+\frac{1}{x^{2}}}$ $dx$
$L$ = $\int_{{\,4}}^{{\,8}}\sqrt{(\frac{x}{4}+\frac{1}{x})^{2}}$ $dx$
$L$ = $\int_{{\,4}}^{{\,8}}(\frac{x}{4}+\frac{1}{x})$ $dx$
$L$ = $\frac{x^{2}}{8}+\ln{x}$$|_{{\,4}}^{{\,8}}$
$L$ = $[(8+\ln{8})-(2+\ln4)]$
$L$ = $6+\ln{2}$
$b)$
$L$ = $\int_{{\,a}}^{{\,b}}\sqrt{1+(\frac{dx}{dy})^{2}}$ $dy$
$\frac{dx}{dy}$ = $\frac{y}{8}-\frac{2}{y}$
$L$ = $\int_{{\,4}}^{{\,12}}\sqrt{1+(\frac{y}{8}-\frac{2}{y})^{2}}$ $dy$
$L$ = $\int_{{\,4}}^{{\,12}}\sqrt{1+\frac{y}{64}^{2}-\frac{1}{2}+\frac{4}{y^{2}}}$ $dy$
$L$ = $\int_{{\,4}}^{{\,12}}\sqrt{\frac{y}{64}^{2}+\frac{1}{2}+\frac{4}{y^{2}}}$ $dy$
$L$ = $\int_{{\,4}}^{{\,12}}\sqrt{(\frac{y}{8}+\frac{2}{y})^{2}}$ $dy$
$L$ = $\int_{{\,4}}^{{\,12}}(\frac{y}{8}+\frac{2}{y})$ $dy$
$L$ = $\frac{y^{2}}{16}+2\ln{y}$$|_{{\,4}}^{{\,12}}$
$L$ = $[(9+2\ln{12})-(1+2\ln4)]$
$L$ = $8+\ln{9}$
