Answer
$a.\quad 2\sinh^{-1}1$
$b.\quad 2\ln(1+\sqrt{2})$
Work Step by Step
$(a)$
Use table: "Integrals leading to inverse hyperbolic functions"
1. $\displaystyle \quad \int\frac{du}{\sqrt{a^{2}+u^{2}}}=\sinh^{-1}(\frac{u}{a})+C, \quad a\gt 0$
Here, $a=1$,
$u(x)=3x.\displaystyle \qquad du=3dx, \quad (dx=\frac{du}{3})$
The integral bounds change to $\left\{\begin{array}{lll}
x=\frac{1}{3} & \rightarrow & u=1\\
x=0 & \rightarrow & u=0
\end{array}\right.$
$\displaystyle \int_{0}^{1/3}\frac{6dx}{\sqrt{1+(3x)^{2}}} =\int_{0}^{1}\frac{2\cdot du}{\sqrt{1^{2}+u^{2}}}$
$=2\displaystyle \cdot \left[ \sinh^{-1}(\frac{u}{1}) \right]_{0}^{1}$
$=2[\sinh^{-1}(1)-\sinh^{-1}0]\qquad $... ( $\sinh 0=0 )$
$=2\sinh^{-1}1$
$(b)$
Using the formulas in the box above these wxercises,
$\sinh^{-1}x=\ln(x+\sqrt{x^{2}+1})$ ,$\quad -\infty \lt x \lt \infty$
$2\sinh^{-1}1=2\ln(1+\sqrt{1+1})$
$=2\ln(1+\sqrt{2})$
