Answer
\[ = \frac{1}{4}\]
Work Step by Step
\[\begin{gathered}
\int_2^\infty {\frac{{dx}}{{\,{{\left( {x + 2} \right)}^2}}}} \, \hfill \\
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finding\,\,the\,\,anti\,derivative \hfill \\
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\int_{}^{} {\frac{{dx}}{{\,{{\left( {x + 2} \right)}^2}}}} \,\, = \,\, - \frac{1}{{x + 2}}\, + C \hfill \\
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use\,\,the\,\,Definition\,\,of\,\,improper\,\,integral\, \hfill \\
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\int_a^\infty {f\,\left( x \right)} \,dx = \,\,\,\,\,\mathop {\,\lim }\limits_{b \to \infty } \int_a^b {f\,\left( x \right)dx} \hfill \\
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\int_2^\infty {\frac{{dx}}{{\,{{\left( {x + 2} \right)}^2}}}} \,\,\, = \,\,\,\,\mathop {\,\lim }\limits_{b \to \infty } \,\int_2^b {\frac{{dx}}{{\,{{\left( {x + 2} \right)}^2}}}dx} \hfill \\
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= \,\,\,\mathop {\,\lim }\limits_{b \to \infty } \,\,\,\,\left[ { - \frac{1}{{x + 2}}} \right]_2^b \hfill \\
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use\,\,the\,\,ftc \hfill \\
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= \,\,\mathop {\,\lim }\limits_{b \to \infty } \,\,\,\,\left[ { - \frac{1}{{b + 2}} + \frac{1}{4}} \right] \hfill \\
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evaluate\,\,the\,\,\lim it \hfill \\
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= \frac{1}{4} \hfill \\
\end{gathered} \]