Answer
$a.\displaystyle \quad\frac{\mathrm{c}\mathrm{s}\mathrm{c}\mathrm{h}^{-1}(\frac{1}{2})-\mathrm{c}\mathrm{s}\mathrm{c}\mathrm{h}^{-1}(1)}{2}$
$b.\displaystyle \quad\frac{1}{2}\ln(\frac{2+\sqrt{5}}{1+\sqrt{2}})$
Work Step by Step
$(a)$
Use table: "Integrals leading to inverse hyperbolic functions"
$\displaystyle \int\frac{du}{u\sqrt{a^{2}+u^{2}}}=-\frac{1}{a}\mathrm{c}\mathrm{s}\mathrm{c}\mathrm{h}^{-1}|\frac{u}{a}|+C,\quad u\neq 0$ and $a\gt 0$
$a=2,$
$u=x, \quad du=dx$
$\displaystyle \int_{1}^{2}\frac{dx}{x\sqrt{1-16x^{2}}}= \left[ -\frac{1}{2}\mathrm{c}\mathrm{s}\mathrm{c}\mathrm{h}^{-1}|\frac{x}{2}| \right]_{1}^{2} $
$=\displaystyle \frac{1}{2}[-\mathrm{c}\mathrm{s}\mathrm{c}\mathrm{h}^{-1}1+\mathrm{c}\mathrm{s}\mathrm{c}\mathrm{h}^{-1}\frac{1}{2}]$
$=\displaystyle \frac{\mathrm{c}\mathrm{s}\mathrm{c}\mathrm{h}^{-1}(\frac{1}{2})-\mathrm{c}\mathrm{s}\mathrm{c}\mathrm{h}^{-1}(1)}{2}$
$(b)$
Using the formulas in the box above these exercises,
$\displaystyle \mathrm{c}\mathrm{s}\mathrm{c}\mathrm{h}^{-1}x=\ln(\frac{1}{x}+\frac{\sqrt{1+x^{2}}}{|x|})$ ,$\quad x\neq 0$
$\displaystyle \frac{\mathrm{c}\mathrm{s}\mathrm{c}\mathrm{h}^{-1}(\frac{1}{2})-\mathrm{c}\mathrm{s}\mathrm{c}\mathrm{h}^{-1}(1)}{2}=\frac{\ln(2+\frac{\sqrt{1+\frac{1}{4}}}{\frac{1}{2}})-\ln(1+\frac{\sqrt{1+1}}{1})}{2}$
$=\displaystyle \frac{\ln(2+2\sqrt{5/4})-\ln(1+\sqrt{2})}{2}$
$=\displaystyle \frac{1}{2}(\ln\frac{2+\frac{2\sqrt{5}}{2}}{1+\sqrt{2}})$
$=\displaystyle \frac{1}{2}\ln(\frac{2+\sqrt{5}}{1+\sqrt{2}})$
