積分(2)の1．(6)の解答

$3$\displaystyle{ \int \frac{\hspace{2}1\hspace{2}}{\hspace{2} {x}^{2}+9 \hspace{2}}dx }$
$3$\displaystyle{\begin{array}{lll}=& \int \frac{\hspace{2}1\hspace{2}}{\hspace{2}9\hspace{2}}\cdot \frac{\hspace{2}1\hspace{2}}{\hspace{2} \frac{\hspace{2} {x}^{2} \hspace{2}}{\hspace{2}9\hspace{2}}+1 \hspace{2}}dx & \vspace{6}\\ =& \frac{\hspace{2}1\hspace{2}}{\hspace{2}9\hspace{2}}\int \frac{\hspace{2}1\hspace{2}}{\hspace{2} { $ \frac{\hspace{2}x\hspace{2}}{\hspace{2}3\hspace{2}} $ }^{2}+1 \hspace{2}} dx & \vspace{6}\\ =& \frac{\hspace{2}1\hspace{2}}{\hspace{2}9\hspace{2}}\cdot 3{ \tan }^{ - 1 }\frac{\hspace{2}x\hspace{2}}{\hspace{2}3\hspace{2}}+C & \vspace{6}\\ =& \frac{\hspace{2}1\hspace{2}}{\hspace{2}3\hspace{2}}{ \tan }^{ - 1 }\frac{\hspace{2}x\hspace{2}}{\hspace{2}3\hspace{2}}+C & \vspace{6}\\ \end{array}}$
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