Answer
\[ = \frac{{\,\left( {x + 5} \right)}}{2}\sqrt {\,{x^2} + 10x} - \frac{{25}}{2}\ln \left| {\,\left( {x + 5} \right) + \sqrt {\,{x^2} + 10x} } \right| + C\]
Work Step by Step
\[\begin{gathered}
\int_{}^{} {\sqrt {{x^2} + 10x} \,dx} \,\, \hfill \\
\hfill \\
complete\,\,the\,square \hfill \\
\hfill \\
= \int_{}^{} {\sqrt {{x^2} + 10x + 25 - 25} } \,dx \hfill \\
\hfill \\
factor \hfill \\
\hfill \\
= \int_{}^{} {\sqrt {\,{{\left( {x + 5} \right)}^2} - 25} } \,dx \hfill \\
\hfill \\
= \int_{}^{} {\sqrt {{u^2} - 25} \,du} \hfill \\
\hfill \\
use\,\,the\,\,integration\,\,formula \hfill \\
\hfill \\
\int_{}^{} {\sqrt {{x^2} - {a^2}} \,dx} = \frac{x}{2}\sqrt {{x^2} - {a^2}} - \frac{{{a^2}}}{2}\ln \left| {x + \sqrt {{x^2} - {a^2}} } \right| + C \hfill \\
\hfill \\
then \hfill \\
\hfill \\
= \int_{}^{} {\sqrt {{u^2} - 25} \,du} = \frac{u}{2}\sqrt {{u^2} - 25} - \frac{{25}}{2}\ln \left| {u + \sqrt {{u^2} - 25} } \right| + C \hfill \\
\hfill \\
substitute\,\,back\,\,u = x + 5 \hfill \\
\hfill \\
= \frac{{\,\left( {x + 5} \right)}}{2}\sqrt {\,{{\left( {x + 5} \right)}^2} - 25} - \frac{{25}}{2}\ln \left| {\,\left( {x + 5} \right) + \sqrt {\,{{\left( {x + 5} \right)}^2} - 25} } \right| + C \hfill \\
\hfill \\
simplify \hfill \\
\hfill \\
= \frac{{\,\left( {x + 5} \right)}}{2}\sqrt {\,{x^2} + 10x} - \frac{{25}}{2}\ln \left| {\,\left( {x + 5} \right) + \sqrt {\,{x^2} + 10x} } \right| + C \hfill \\
\hfill \\
\end{gathered} \]
