Answer
$a.\displaystyle \quad-{\rm sech}^{-1}(\frac{12}{13})+{\rm sech}^{-1}(\frac{4}{5})$
$b.\displaystyle \quad\ln\frac{4}{3}$
Work Step by Step
$(a)$
Use table: "Integrals leading to inverse hyperbolic functions"
$\displaystyle \int\frac{du}{u\sqrt{a^{2}-u^{2}}}=-\frac{1}{a}{\rm sech}^{-1}(\frac{u}{a})+C, \quad 0 \lt u \lt a$
$a=1,$
$u=4x, \displaystyle \quad (x=\frac{u}{4}),\quad dx=\frac{dx}{4}$
Bounds change :$\left\{\begin{array}{lll}
x=3/13 & \rightarrow & u=12/13\\
x=1/5 & \rightarrow & u=4/5
\end{array}\right.$
$\displaystyle \int_{1/5}^{3/13}\frac{dx}{x\sqrt{1-16x^{2}}}=\int_{4/5}^{12/13}\frac{\frac{du}{4}}{\frac{u}{4}\sqrt{1-u^{2}}}$
$=\displaystyle \int_{4/5}^{12/13}\frac{du}{u\sqrt{1-u^{2}}}$
$= \left[ -{\rm sech}^{-1}u \right]_{4/5}^{12/13} $
$=-{\rm sech}^{-1}(\displaystyle \frac{12}{13})+{\rm sech}^{-1}(\frac{4}{5})$
$(b)$
Using the formulas in the box above these exercises,
${\rm sech}^{-1}x=\displaystyle \ln(\frac{1+\sqrt{1-x^{2}}}{x})$ ,$\quad 0 \lt x\leq 1$
$-{\rm sech}^{-1}(\displaystyle \frac{12}{13})+{\rm sech}^{-1}(\frac{4}{5})=-\ln(\frac{1+\sqrt{1-(\frac{144}{169})}}{\frac{12}{13}})+\ln(\frac{1+\sqrt{1-(\frac{16}{25})}}{\frac{4}{5}})$
$=-\displaystyle \ln(\frac{1+\sqrt{\frac{25}{169}}}{\frac{12}{13}})+\ln(\frac{1+\sqrt{\frac{9}{25}}}{\frac{4}{5}})$
$=-\displaystyle \ln(\frac{1+\frac{5}{13}}{\frac{12}{13}})+\ln(\frac{1+\frac{3}{5}}{\frac{4}{5}})$
$=-\displaystyle \ln(\frac{\frac{18}{13}}{\frac{12}{13}})+\ln(\frac{\frac{8}{5}}{\frac{4}{5}})$
$=-\displaystyle \ln(\frac{18}{12})+\ln(\frac{8}{4})$
$=\ln(2)-\ln(1.5)\qquad $... ( $\displaystyle \ln a-\ln b=\ln\frac{a}{b}$ )
$=\displaystyle \ln\frac{2}{1.5}$
$=\displaystyle \ln\frac{4}{3}$
