## Calculus: Early Transcendentals (2nd Edition)

$= \frac{{\,\left( {x - 4} \right)}}{2}\sqrt {\,{x^2} - 8x} - 8\ln \left| {\,\left( {x - 4} \right) + \sqrt {\,{x^2} - 8x} } \right| + C$
$\begin{gathered} \int_{}^{} {\sqrt {{x^2} - 8x} \,dx} \hfill \\ \hfill \\ complete\,\,the\,\,square \hfill \\ \hfill \\ = \int_{}^{} {\sqrt {{x^2} - 8x + 16 - 16} \,\,dx} \hfill \\ \hfill \\ factor \hfill \\ \hfill \\ = \int_{}^{} {\sqrt {\,{{\left( {x - 4} \right)}^2} - 16} dx} \hfill \\ \hfill \\ = \int_{}^{} {\sqrt {{u^2} - 16} \,du} \hfill \\ \hfill \\ use\,\,the\,\,integration\,\,formula \hfill \\ \hfill \\ \int_{}^{} {\sqrt {{x^2} - {a^2}} \,dx} = \frac{x}{2}\sqrt {{x^2} - {a^2}} - \frac{{{a^2}}}{2}\ln \left| {x + \sqrt {{x^2} - {a^2}} } \right| + C \hfill \\ \hfill \\ \int_{}^{} {\sqrt {{u^2} - 16} \,du} = \frac{u}{2}\sqrt {{u^2} - 16} - \frac{{16}}{2}\ln \left| {u + \sqrt {{u^2} - 16} } \right| + C \hfill \\ \hfill \\ substitute\,\,back\,\,u = x + 5 \hfill \\ \hfill \\ = \frac{{\,\left( {x - 4} \right)}}{2}\sqrt {\,{{\left( {x - 4} \right)}^2} - 16} - 8\ln \left| {\,\left( {x - 4} \right) + \sqrt {\,{{\left( {x - 4} \right)}^2} - 16} } \right| + C \hfill \\ \hfill \\ Simplify \hfill \\ \hfill \\ = \frac{{\,\left( {x - 4} \right)}}{2}\sqrt {\,{x^2} - 8x} - 8\ln \left| {\,\left( {x - 4} \right) + \sqrt {\,{x^2} - 8x} } \right| + C \hfill \\ \hfill \\ \hfill \\ \end{gathered}$