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