## Calculus: Early Transcendentals (2nd Edition)

$= - \frac{{\ln x}}{{9{x^9}}} - \frac{1}{{81{x^9}}} + C$
$\begin{gathered} \int_{}^{} {\frac{{\ln x}}{{{x^{10}}}}\,dx} \hfill \\ rewrite \hfill \\ \hfill \\ = \int_{}^{} {\,\left( {\ln \,x} \right)\,\left( {{x^{ - 10}}} \right)\,dx} \hfill \\ \hfill \\ set\,\,\,the\,\,substitution \hfill \\ \hfill \\ u = \ln x\,\,\,\,\,\,\,\, \to \,\,\,\,du = \frac{1}{x}dx \hfill \\ dv = {x^{10}}dx\, \to \,\,\,v = - \frac{{ - {x^{ - 9}}}}{9} \hfill \\ \hfill \\ use\,\,uv - \int_{}^{} {vdu} \hfill \\ \hfill \\ {\text{replacing the values }}{\text{in the equation}} \hfill \\ \hfill \\ = \,\left( {\ln x} \right)\,\left( {\frac{{ - {x^{ - 9}}}}{9}} \right) - \int_{}^{} {\,\left( { - \frac{{{x^{ - 9}}}}{9}} \right)\,\left( {\frac{1}{x}} \right)dx} \hfill \\ \hfill \\ simplify \hfill \\ \hfill \\ = - \frac{{\ln x}}{{9{x^9}}} + \frac{1}{9}\int_{}^{} {\frac{1}{{{x^{10}}}}dx} \hfill \\ \hfill \\ or \hfill \\ \hfill \\ = - \frac{{\ln x}}{{9{x^9}}} + \frac{1}{9}\int_{}^{} {{x^{ - 10}}dx} \hfill \\ \hfill \\ integrate\,\,{\text{using }}\int {{x^n}dx = \frac{{{x^{n + 1}}}}{{n + 1}} + C} \hfill \\ \hfill \\ = - \frac{{\ln \,x}}{{9{x^9}}} + \frac{1}{9}\,\left( {\frac{{ - 1}}{{9{x^4}}}} \right) + C \hfill \\ \hfill \\ simplify \hfill \\ \hfill \\ = - \frac{{\ln x}}{{9{x^9}}} - \frac{1}{{81{x^9}}} + C \hfill \\ \end{gathered}$