How to integrate $\frac{\sqrt{z+1}}{z}$ How to integrate $\frac{\sqrt{z+1}}{z}$
Anyone could help me? Thanks

$u=\sqrt{z+1}$, $z=u^2-1$. Then
$$\begin{gathered}
  \int {\frac{{\sqrt {z + 1} }}
{z}dz}  = \int {\frac{{2{u^2}}}
{{{u^2} - 1}}du}  = 2\int {\left( {1 + \frac{1}
{{{u^2} - 1}}} \right)du}  = 2\int {\left[ {1 + \frac{1}
{2}\left( {\frac{1}
{{u - 1}} - \frac{1}
{{u + 1}}} \right)} \right]du}  \hfill \\
   = 2\left( {u + \frac{1}
{2}\ln \left| {\frac{{u - 1}}
{{u + 1}}} \right| + C} \right) = 2\left( {\sqrt {z + 1}  + \frac{1}
{2}\ln \left| {\frac{{\sqrt {z + 1}  - 1}}
{{\sqrt {z + 1}  + 1}}} \right| + C} \right) \hfill \\ 
\end{gathered} $$
