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+%%% fig.tex 2016-4-25 8:55
+%%% fig.sce 2016-4-25 8:55
+{\unitlength=8mm%
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+%
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+\settoheight{\Height}{部分分数への分解 $\dfrac{x^3}{\left(x+1\right)\,\left(x+2\right)}=\dfrac{8}{x+2}-\dfrac{1}{x+1}+x-3$}\settodepth{\Depth}{部分分数への分解 $\dfrac{x^3}{\left(x+1\right)\,\left(x+2\right)}=\dfrac{8}{x+2}-\dfrac{1}{x+1}+x-3$}\setlength{\Height}{-0.5\Height}\setlength{\Depth}{0.5\Depth}\addtolength{\Height}{\Depth}%
+\put(0.0500,3.0000){\hspace*{\Width}\raisebox{\Height}{部分分数への分解 $\dfrac{x^3}{\left(x+1\right)\,\left(x+2\right)}=\dfrac{8}{x+2}-\dfrac{1}{x+1}+x-3$}}%
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