積分(4)の1．(2)の解答

$3$\displaystyle{ \int \frac{\hspace{2}1\hspace{2}}{\hspace{2} $x- 2$$x+3$ \hspace{2}}dx }$
$3$\displaystyle{ \frac{\hspace{2}a\hspace{2}}{\hspace{2} x- 2 \hspace{2}}+\frac{\hspace{2}b\hspace{2}}{\hspace{2} x+3 \hspace{2}} }$
$3$\displaystyle{\begin{array}{lll}=& \frac{\hspace{2} a $ x+3 $ +b $ x- 2 $ \hspace{2}}{\hspace{2} $ x- 2 $ $ x+3 $ \hspace{2}} & \vspace{6}\\ =& \frac{\hspace{2} x $ a+b $ +3a- 2b \hspace{2}}{\hspace{2} $ x- 2 $ $ x+3 $ \hspace{2}} & \vspace{6}\\ \end{array}}$
係数を比較すると
$3$\displaystyle{a+b=0}$ と $3$\displaystyle{3a- 2b=1}$ より
$3$\displaystyle{a=\frac{\hspace{2}1\hspace{2}}{\hspace{2}5\hspace{2}}}$ $3$\displaystyle{b=- \frac{\hspace{2}1\hspace{2}}{\hspace{2}5\hspace{2}}}$
与式は以下のとおりになる $3$\displaystyle{\begin{array}{lll}=& \int \frac{\hspace{2}1\hspace{2}}{\hspace{2}5\hspace{2}} $ \frac{\hspace{2}1\hspace{2}}{\hspace{2} x- 2 \hspace{2}}- \frac{\hspace{2}1\hspace{2}}{\hspace{2} x+3 \hspace{2}} $ dx & \vspace{6}\\ =& \frac{\hspace{2}1\hspace{2}}{\hspace{2}5\hspace{2}} $ \log \| x- 2 \| - \log \| x+3 \| $ +C & \vspace{6}\\ =& \frac{\hspace{2}1\hspace{2}}{\hspace{2}5\hspace{2}}\log \| \frac{\hspace{2} x- 2 \hspace{2}}{\hspace{2} x+3 \hspace{2}} \| +C & \vspace{6}\\ \end{array}}$

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