積分(1)の1．(9)の解答

$3$\displaystyle{ \int \frac{\hspace{2} {x}^{3}- 1 \hspace{2}}{\hspace{2} x- 1 \hspace{2}} dx }$
$3$\displaystyle{ \begin{array}{lll}=& \int \frac{\hspace{2} $ x- 1 $ $ {x}^{2}+x+1 $ \hspace{2}}{\hspace{2} x- 1 \hspace{2}} & \vspace{6}\\ =& \int $ {x}^{2}+x+1 $ dx & \vspace{6}\\ =& \frac{\hspace{2}1\hspace{2}}{\hspace{2}3\hspace{2}}{x}^{3}+\frac{\hspace{2}1\hspace{2}}{\hspace{2}2\hspace{2}}{x}^{2}+x+C & \vspace{6}\\ \end{array} }$
　 $3$C$ は積分定数

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