next up previous
次へ: 部分積分 上へ: 戻る: 置換積分

部分分数展開

  1. $\displaystyle I=\int \frac{x^2+x+1}{x+2}dx
=\int\left\{x-1+\frac{3}{x+2}\right\}dx$
  2. $\displaystyle I=\int \frac{1}{(x+3)(x+4)}dx
=\int\left\{\frac{A}{x+3}+\frac{B}{x+4}\right\}dx$

    \begin{eqnarray*}
A&=& \left.\frac{1}{x+4}\right\vert _{x=-3}=1\\
B&=& \left.\frac{1}{x+3}\right\vert _{x=-4}=-1
\end{eqnarray*}

  3. $\displaystyle I=\int \frac{x-7}{x^2-2x-3}dx
=\int\left\{\frac{A}{x+1}+\frac{B}{x-3}\right\}dx$

    \begin{eqnarray*}
A&=& \left.\frac{x-7}{x-3}\right\vert _{x=-1}\\
B&=& \left.\frac{x-7}{x+1}\right\vert _{x=3}
\end{eqnarray*}

  4. $\displaystyle I=\int \frac{2x+3}{x(x+2)(x+3)}dx$

    \begin{displaymath}
\frac{2x+3}{x(x+2)(x+3)}
=\frac{a}{x}+\frac{b}{x+2}+\frac{c}{x+3}
\end{displaymath}

    \begin{eqnarray*}
a&=& \left.\frac{2x+3}{(x+2)(x+3)}\right\vert _{x=0}\\
b&=&...
...t _{x=-2}\\
c&=& \left.\frac{2x+3}{x(x+2)}\right\vert _{x=-3}
\end{eqnarray*}



T.Kinoshita 平成15年10月21日