Answer
$a.\displaystyle \quad\tanh^{-1}(\frac{1}{2})$
$b.\displaystyle \quad\frac{1}{2}\ln 3$
Work Step by Step
$(a)$
Use table: "Integrals leading to inverse hyperbolic functions"
3. $\displaystyle \int\frac{du}{a^{2}-u^{2}}=\left\{\begin{array}{ll}
\frac{1}{a}\tanh^{-1}(\frac{u}{a})+C, & u^{2} \lt a^{2}\\
\frac{1}{a}\coth^{-1}(\frac{u}{a})+C, & u^{2}\gt a^{2}
\end{array}\right.$
on the interval $[0,1/2]$, the graph of $y=x^{2}$ is below the graph of $y=x$, so we apply the $\tanh^{-1}$ case.
$a=1, u(x)=x, du=dx,$
$\displaystyle \int_{0}^{1}\frac{dx}{1-x^{2}}= \left[ \tanh^{-1}(\frac{x}{1}) \right]_{0}^{1/2}$
$= \displaystyle \tanh^{-1}(\frac{1}{2}) - \tanh^{-1}(0)$
$= \displaystyle \tanh^{-1}(\frac{1}{2})$
$(b)$
Using the formulas in the box above these exercises,
$\displaystyle \tanh^{-1}x=\frac{1}{2}\ln \displaystyle \frac{1+x}{1-x},\quad |x| \lt 1$
$\displaystyle \tanh^{-1}(\frac{1}{2})=\frac{1}{2}\ln \displaystyle \frac{1+\frac{1}{2}}{1-\frac{1}{2}}$
$=\displaystyle \frac{1}{2}\ln \displaystyle \frac{\frac{3}{2}}{\frac{1}{2}}$
$=\displaystyle \frac{1}{2}\ln 3$
