Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 7 - Integration Techniques - 7.6 Other Integration Strategies - 7.6 Exercises: 23

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

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