Answer
\[ = \ln 2\]
Work Step by Step
\[\begin{gathered}
\int_1^\infty {\frac{{dv}}{{v\,\left( {v + 1} \right)}}} \hfill \\
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integrating\,\,by\,\,partial\,\,fractions \hfill \\
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\int_{}^{} {\frac{{dv}}{{v\,\left( {v + 1} \right)}}} \,\, = \,\,\int_{}^{} {\,\left( {\frac{1}{v} - \frac{1}{{v + 1}}} \right)dv} \hfill \\
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= \ln \left| v \right| - \ln \left| {v + 1} \right| + C \hfill \\
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\log \,\,property \hfill \\
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= \ln \left| {\frac{v}{{v + 1}}} \right| + C \hfill \\
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use\,\,Definition\,\,of\,\,improper\,\,integral\,\, \hfill \\
\int_a^\infty {f\,\left( x \right)} \,dx = \,\,\,\,\,\mathop {\,\lim }\limits_{b \to \infty } \int_0^b {f\,\left( x \right)dx} \hfill \\
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then \hfill \\
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\int_1^\infty {\frac{{dv}}{{v\,\left( {v + 1} \right)}}} = \mathop {\lim }\limits_{b \to \infty } \int_a^b {\frac{{dv}}{{v\,\left( {v + 1} \right)}}} \hfill \\
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= \mathop {\lim }\limits_{b \to \infty } \,\,\ln \,\left| {\frac{v}{{v + 1}}} \right|_1^b \hfill \\
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use\,\,the\,\,ftc \hfill \\
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= \mathop {\lim }\limits_{b \to \infty } \,\left( {\ln \left| {\frac{b}{{b + 1}}} \right| - \ln \left| {\frac{1}{{1 + 1}}} \right|} \right) \hfill \\
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= \mathop {\lim }\limits_{b \to \infty } \,\left( {\ln \,\left( {1 - \frac{1}{{b + 1}}} \right) - \ln \frac{1}{2}} \right) \hfill \\
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{\text{Simplify}} \hfill \\
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= \,\left( {\ln \,\left( {1 - 0} \right) + \ln 2} \right) = \ln 2 \hfill \\
\end{gathered} \]