## Calculus: Early Transcendentals (2nd Edition)

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