Answer
\[ = x - \ln \left| {x + 1} \right| + C\]
Work Step by Step
\[\begin{gathered}
\int_{}^{} {\frac{{dx}}{{{x^{ - 1}} + 1}} = \int_{}^{} {\frac{{xdx}}{{1 + x}}} } \hfill \\
\hfill \\
or \hfill \\
\hfill \\
= \int_{}^{} {\frac{{\,\left( {x + 1} \right) - 1}}{{x + 1}}} dx \hfill \\
\hfill \\
split\,\,the\,\,{\text{integrand}} \hfill \\
\hfill \\
= \int_{}^{} {\,\left( {1 - \frac{1}{{x + 1}}} \right)} dx \hfill \\
\hfill \\
\int_{}^{} {x - \int_{}^{} {\frac{1}{{x + 1}}dx + } } C \hfill \\
\hfill \\
integrate \hfill \\
\hfill \\
= x - \ln \left| {x + 1} \right| + C \hfill \\
\end{gathered} \]
