Answer
\[ = \frac{1}{{1458}}{\tan ^{ - 1}}\,\left( {\frac{x}{9}} \right) + \frac{x}{{162\,\left( {81 + {x^2}} \right)}} + C\]
Work Step by Step
\[\begin{gathered}
\int_{}^{} {\frac{{dx}}{{\,{{\left( {81 + {x^2}} \right)}^2}}}} \hfill \\
\hfill \\
rewrite,\,\,use\,\,\sqrt[n]{{{a^m}}} = {a^{m/n}} \hfill \\
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= \int_{}^{} {\frac{{dx}}{{\,{{\left( {\sqrt {{x^2} + {9^2}} } \right)}^4}}}} \hfill \\
\hfill \\
the\,\,\,integral\,contains\,\,the\,\,form\,\,{x^2} - {a^2}\,\, \hfill \\
\hfill \\
x = 9\tan \theta \,\,\,\,\,\, \to \,\,\,\,\,dx = 9{\sec ^2}\theta d\theta \hfill \\
and\,\,\sqrt {{x^2} + 81} = 9\sec \theta \hfill \\
\hfill \\
apply\,\,the\,\,substitution \hfill \\
\hfill \\
= \int_{}^{} {\frac{{dx}}{{\,\left( {\sqrt {{x^2} + {9^2}} } \right)}}} = \int_{}^{} {\frac{{9{{\sec }^2}\theta d\theta }}{{{9^4}{{\sec }^4}\theta }}} \hfill \\
\hfill \\
simplify \hfill \\
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= \frac{1}{{{9^3}}}\int_{}^{} {\frac{1}{{\sec \theta }}} d\theta \hfill \\
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= \frac{1}{{729}}\,\int_{}^{} {{{\cos }^2}\theta d\theta } = \frac{1}{{729}}\int_{}^{} {\,\left( {\frac{{1 + \cos 2\theta }}{2}} \right)d\theta } \hfill \\
\hfill \\
integrate \hfill \\
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= \frac{1}{{729}}\,\left( {\frac{\theta }{2} + \frac{{\sin 2\theta }}{4}} \right) + C \hfill \\
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= \frac{1}{{729}}\,\left( {\frac{\theta }{2} + \frac{{\sin \theta \cos \theta }}{2}} \right) + C \hfill \\
\hfill \\
substitute\,\,for\,\,\sin \theta {\text{ and cos}}\theta \hfill \\
\hfill \\
= \frac{1}{{1458}}{\tan ^{ - 1}}\,\left( {\frac{x}{9}} \right) + \frac{1}{{1458}}\,\left( {\frac{x}{{\sqrt {81 + 2} }}} \right)\,\left( {\frac{9}{{\sqrt {81 + {x^2}} }}} \right) + C \hfill \\
\hfill \\
= \frac{1}{{1458}}{\tan ^{ - 1}}\,\left( {\frac{x}{9}} \right) + \frac{x}{{162\,\left( {81 + {x^2}} \right)}} + C \hfill \\
\end{gathered} \]
