Calculus: Early Transcendentals (2nd Edition)

$= \,\frac{2}{{\sqrt a }}\ln \left| {\frac{{\sqrt y }}{{\sqrt y - \sqrt a }}} \right| + C$
$\begin{gathered} \int_{}^{} {\frac{{dy}}{{y\,\left( {\sqrt a - \sqrt y } \right)}}\,\,,\,\,\,\,for\,\,a > 0} \hfill \\ \hfill \\ {\text{integrand}} \hfill \\ \hfill \\ \frac{1}{{y\,\left( {\sqrt a - \sqrt y } \right)}} \hfill \\ \hfill \\ {\text{Apply partial fraction decomposition}} \hfill \\ \hfill \\ \frac{1}{{y\,\left( {\sqrt u - \sqrt y } \right)}} = \frac{1}{{\,{{\left( {\sqrt y } \right)}^2}\,\left( {\sqrt a - \sqrt y } \right)}} \hfill \\ \hfill \\ therefore \hfill \\ \hfill \\ = \frac{1}{{\,\left( {\sqrt y } \right)\,\left( {\sqrt {a - \sqrt y } } \right)}}\frac{1}{{\sqrt y }} = \,\left( {\frac{1}{{\sqrt a \sqrt y }} - \frac{1}{{\sqrt a \,\left( {\sqrt y - \sqrt a } \right)}}} \right)\frac{1}{{\sqrt y }} \hfill \\ \hfill \\ integrating \hfill \\ \hfill \\ \int_{}^{} {\frac{{dy}}{{y\,\left( {\sqrt a - \sqrt y } \right)}}\,\, = } \int_{}^{} {\,\left( {\frac{1}{{\sqrt a \sqrt y }} - \frac{1}{{\sqrt a \,\left( {\sqrt y - \sqrt a } \right)}}} \right)} \frac{1}{{\sqrt y }}dy \hfill \\ \hfill \\ set\,\,\,u = \sqrt y \hfill \\ \hfill \\ apply\,\,the\,\,substitution \hfill \\ \hfill \\ = \int_{}^{} {\,\left( {\frac{1}{{\sqrt a u}} - \frac{1}{{\sqrt a \,\left( {u - \sqrt a } \right)}}} \right)} \,\left( {2du} \right) \hfill \\ {\text{integrate}} \hfill \\ \hfill \\ = \frac{2}{{\sqrt a }}\,\left( {\ln \left| u \right| - \ln \left| {u - \sqrt a } \right|} \right) + C \hfill \\ \hfill \\ substitute\,\,for\,\,\,u \hfill \\ \hfill \\ = \frac{2}{{\sqrt a }}\,\left( {\ln \left| {\frac{{\sqrt y }}{{\sqrt y - \sqrt a }}} \right|} \right) + C \hfill \\ \hfill \\ or \hfill \\ \hfill \\ = \,\frac{2}{{\sqrt a }}\ln \left| {\frac{{\sqrt y }}{{\sqrt y - \sqrt a }}} \right| + C \hfill \\ \end{gathered}$