Answer
\[\, = \ln \left| {\frac{{\sqrt {1 + 2x} - 1}}{{\sqrt {1 + 2x }+1 }}} \right| + C\]
Work Step by Step
\[\begin{gathered}
\int_{}^{} {\frac{{dx}}{{x\sqrt {1 + 2x} }}} \hfill \\
\hfill \\
set{\text{ }}1 + 2x = {u^2} \hfill \\
\hfill \\
solve\,\,for\,\,x \hfill \\
\hfill \\
x = \frac{{{u^2} - 1}}{2}\,\,\,\,\,\,then\,\,\,\,dx = udu \hfill \\
\hfill \\
therefore \hfill \\
\hfill \\
\int_{}^{} {\frac{{dx}}{{x\sqrt {1 + 2x} }}} = \int_{}^{} {\frac{{du}}{{{u^2} - 1}}} \hfill \\
\hfill \\
Applying\,\,partial\,fractions \hfill \\
\frac{1}{{{u^2} - 1}} = \frac{A}{{u + 1}} + \frac{B}{{u - 1}} \hfill \\
\hfill \\
A = - 1,{\text{ and}}\,\,B = 1 \hfill \\
\hfill \\
\frac{1}{{{u^2} - 1}} = - \frac{1}{{u + 1}} + \frac{1}{{u - 1}} \hfill \\
\hfill \\
\int_{}^{} {\frac{{du}}{{{u^2} - 1}}} = \int_{}^{} {\,\left( {\frac{1}{{u - 1}} - \frac{1}{{u + 1}}} \right)\,du} \hfill \\
\hfill \\
{\text{integrate}} \hfill \\
\hfill \\
\ln \left| {u - 1} \right| - \ln \left| {u + 1} \right| \hfill \\
\hfill \\
{\text{property}}\,\,{\text{of}}\,\,{\text{logarithms}} \hfill \\
\hfill \\
= \ln \left| {\frac{{u - 1}}{{u + 1}}} \right| + C\, \hfill \\
\hfill \\
substitute\,\,for\,\,\,u \hfill \\
\hfill \\
\, = \ln \left| {\frac{{\sqrt {1 + 2x} - 1}}{{\sqrt {1 + 2x }+1 }}} \right| + C \hfill \\
\hfill \\
\end{gathered} \]
