Answer
$$\tanh ^{-1} \frac{1}{2}-\tanh ^{-1} \frac{1}{3}$$
Work Step by Step
Since
$$\int \frac{d x}{a^2-x^{2}}=\frac{1}{a}\tanh^{-1}(x/a) $$
Then
\begin{align*}
\int_{1 / 3}^{1 / 2} \frac{d x}{1-x^{2}}&=\tanh ^{-1} x\bigg|_{1 / 3} ^{1 / 2}\\
&=\tanh ^{-1} \frac{1}{2}-\tanh ^{-1} \frac{1}{3}
\end{align*}
