積分(2)の1．(6)のヒント

$3$\displaystyle{ \int \frac{\hspace{2}1\hspace{2}}{\hspace{2} {x}^{2}+9 \hspace{2}}dx }$

$3$\displaystyle{\displaystyle{ \int \frac{\hspace{2}1\hspace{2}}{\hspace{2} 1+{x}^{2} \hspace{2}}dx }={\tan }^{ - 1 }x+C}$

を使うために

$3$\displaystyle{\begin{array}{lll} \displaystyle{ \int \frac{\hspace{2}1\hspace{2}}{\hspace{2} {x}^{2}+9 \hspace{2}}dx } & =\displaystyle{ \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}}\displaystyle{ \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}\\ \end{array}}$

のように式を変形する。この後は，次の公式と連携させて積分する。

　 $3$\displaystyle{\begin{array}{lll}\displaystyle{ \int f $ ax+b $ dx }& \vspace{6}\\ \text{\hspace{5}}=\frac{\hspace{2}1\hspace{2}}{\hspace{2}a\hspace{2}}F $ ax+b $ +C& \vspace{6}\\ \end{array}}$ 　

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