Answer
\[ = \frac{1}{{\,\left( {p - 1} \right)\,{{\left( {\ln 2} \right)}^{p - 1}}}}\]
Work Step by Step
\[\begin{gathered}
\int_2^\infty {\frac{{dx}}{{x{{\ln }^p}x}}} \hfill \\
\hfill \\
use\,\,the\,\,definition\,\,of\,\,improper\,\,{\text{integrals}} \hfill \\
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\int_2^\infty {\frac{{dx}}{{x{{\ln }^p}x}}} = \mathop {\lim }\limits_{b \to \infty } \int_2^b {\frac{{dx}}{{x{{\ln }^p}x}}} \hfill \\
\hfill \\
set \hfill \\
\ln x = u\,\,\,then\,\,\,\frac{{dx}}{x} = du \hfill \\
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{\text{substituting}}\,\,{\text{and}}\,\,{\text{integrating}} \hfill \\
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\mathop {\lim }\limits_{b \to \infty } \int_2^b {\frac{{dx}}{{x{{\ln }^p}x}}} = \mathop {\lim }\limits_{b \to \infty } d\,\,\left[ {\frac{1}{{\,\left( {1 - p} \right){{\ln }^{p - 1}}x}}} \right]_2^b \hfill \\
\hfill \\
use\,\,the\,\,ftc \hfill \\
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= \mathop {\lim }\limits_{b \to \infty } \,\,\left[ {\frac{1}{{\,\left( {1 - p} \right){{\ln }^{p - 1}}b}} - \frac{1}{{\,\left( {1 - p} \right){{\ln }^{p - 1}}2}}} \right] \hfill \\
\hfill \\
evaluate\,\,the\,\,\lim it \hfill \\
\hfill \\
= \frac{1}{{\,\left( {p - 1} \right)\,{{\left( {\ln 2} \right)}^{p - 1}}}} \hfill \\
\end{gathered} \]
