Answer
\[ = \ln \left| x \right| + C\]
Work Step by Step
\[\begin{gathered}
The\,\,restriction\,\,it\,\,is\,\,necessary\,\, \hfill \\
Because\,,\,\,for\,\,n = - 1 \hfill \\
\frac{{{x^{n + 1}}}}{{n + 1}} = \frac{{{x^{ - 1 + 1}}}}{{ - 1 + 1}} = \frac{{{x^0}}}{0} = \frac{1}{0} \hfill \\
and \hfill \\
\int_{}^{} {{x^{ - 1}}dx} = \int_{}^{} {\frac{1}{x}dx} = \ln \left| x \right| + C \hfill \\
\end{gathered} \]
