Originally Posted by
DeMath $\displaystyle \int {\frac{s}
{{\sqrt {{s^2} + {4^2}} }}ds} = \frac{1}
{2}\int {\frac{{2s}}
{{\sqrt {{s^2} + 16} }}ds} =$$\displaystyle \left\{ \begin{gathered}
\sqrt {{s^2} + 4} = t, \hfill \\
{s^2} = {t^2} - 16, \hfill \\
2sds = 2tdt \hfill \\
\end{gathered} \right\} =$
$\displaystyle = \frac{1}
{2}\int {\frac{{2t}}
{t}dt} = \int {dt} = t + C = \sqrt {{s^2} + 16} + C.$
$\displaystyle \int\limits_0^3 {\frac{s}
{{\sqrt {{s^2} + 16} }}ds = } \left. {\sqrt {{s^2} + 16} } \right|_0^3 = \sqrt {25} - \sqrt {16} = 5 - 4 = 1.$