【微分積分】4-4-1 有理関数の積分｜要点まとめ
このページでは、大学数学の微分積分で扱う「有理関数の積分」について整理します。有理関数の定義や基本的な分解方法を確認し、積分計算の手順を例題を通して理解します。部分分数分解や置換法を活用して、有理関数の積分を効率的に解く方法を身につけましょう。
有理関数とは？基本的な定義と特徴
【有理関数の定義】
実数係数の多項式\(p(x)\)と\(q(x)\)を用いて
\(\displaystyle f(x)=\frac{p(x)}{q(x)}\)
の形に表せる関数\(f(x)\)を有理関数という。
【例題】次の不定積分を求めなさい。
(1)\(\displaystyle \int\frac{x^4+x^3+x^2+x+1}{x^2}dx\)
\(\displaystyle =\int\left(x^2+x+1+\frac{1}{x}+\frac{1}{x^2}\right)dx\)
\(\displaystyle =\frac{x^3}{3}+\frac{x^2}{2}+x+\log|x|-\frac{1}{x}+C\)
\(\displaystyle =\frac{x^3}{3}+\frac{x^2}{2}+x+\log|x|-\frac{1}{x}+C\)
(2)\(\displaystyle \int\frac{x^3}{x+1}dx\)
\(\displaystyle =\int\left(x^2-x+1-\frac{1}{x+1}\right)dx\)
\(\displaystyle =\frac{x^3}{3}-\frac{x^2}{2}+x-\log|x+1|+C\)
\(\displaystyle =\frac{x^3}{3}-\frac{x^2}{2}+x-\log|x+1|+C\)
(3)\(\displaystyle \int\frac{3x^3+4x^2+9x+14}{x^2+3}dx\)
\(\displaystyle =\int\left(3x+4+\frac{2}{x^2+3}\right)dx\)
\(\displaystyle =\frac{3}{2}x^2+4x+\frac{2}{\sqrt{3}}\tan^{-1}\frac{x}{\sqrt{3}}+C\)
\(\displaystyle =\frac{3}{2}x^2+4x+\frac{2}{\sqrt{3}}\tan^{-1}\frac{x}{\sqrt{3}}+C\)
(4)\(\displaystyle \int\frac{2x^3-3x^2+6x+1}{x^2-2x+3}dx\)
\(\displaystyle =\int\left(2x+1+\frac{2x-2}{x^2-2x+3}\right)dx\)
\(\displaystyle =\int\left(2x+1+\frac{(x^2-2x+3)'}{x^2-2x+3}\right)dx\)
\(\displaystyle =x^2+x+\log|x^2-2x+3|+C\)
\(\displaystyle =\int\left(2x+1+\frac{(x^2-2x+3)'}{x^2-2x+3}\right)dx\)
\(\displaystyle =x^2+x+\log|x^2-2x+3|+C\)
(5)\(\displaystyle \int\frac{1}{x^2+4x+7}dx\)
\(\displaystyle =\int\frac{1}{(x+2)^2+3}dx\)
\(\displaystyle =\frac{1}{\sqrt{3}}\tan^{-1}\frac{x+2}{\sqrt{3}}+C\)
\(\displaystyle =\frac{1}{\sqrt{3}}\tan^{-1}\frac{x+2}{\sqrt{3}}+C\)
(6)\(\displaystyle \int\frac{2x+3}{x^2-2x+4}dx\)
\(\displaystyle =\int\left(\frac{2x-2}{x^2-2x+4}+\frac{5}{x^2-2x+4}\right)dx\)
\(\displaystyle =\int\frac{(x^2-2x+4)'}{x^2-2x+4}dx+5\int\frac{1}{(x-1)^2+3}dx\)
\(\displaystyle =\log|x^2-2x+4|+\frac{5}{\sqrt{3}}\tan^{-1}\frac{x-1}{\sqrt{3}}+C\)
\(\displaystyle =\int\frac{(x^2-2x+4)'}{x^2-2x+4}dx+5\int\frac{1}{(x-1)^2+3}dx\)
\(\displaystyle =\log|x^2-2x+4|+\frac{5}{\sqrt{3}}\tan^{-1}\frac{x-1}{\sqrt{3}}+C\)
(7)\(\displaystyle \int\frac{x+3}{x^2+2x+5}dx\)
\(\displaystyle =\int\left(\frac{1}{2}・\frac{2x+2}{x^2+2x+5}+\frac{2}{x^2+2x+5}\right)dx\)
\(\displaystyle =\frac{1}{2}\int\frac{(x^2+2x+5)'}{x^2+2x+5}dx\)
\(\displaystyle \ \ \ +2\int\frac{1}{(x+1)^2+4}dx\)
\(\displaystyle =\frac{1}{2}\log|x^2+2x+5|+\tan^{-1}\frac{x+1}{2}+C\)
\(\displaystyle =\frac{1}{2}\int\frac{(x^2+2x+5)'}{x^2+2x+5}dx\)
\(\displaystyle \ \ \ +2\int\frac{1}{(x+1)^2+4}dx\)
\(\displaystyle =\frac{1}{2}\log|x^2+2x+5|+\tan^{-1}\frac{x+1}{2}+C\)
(8)\(\displaystyle \int\frac{x}{(x^2+4)^2}dx\)
\(\displaystyle =\frac{1}{2}\int\frac{2x}{(x^2+4)^2}dx\)
\(\displaystyle =\frac{1}{2}\int\frac{(x^2+4)'}{(x^2+4)^2}dx\)
\(\displaystyle =-\frac{1}{2(x^2+4)}+C\)
\(\displaystyle =\frac{1}{2}\int\frac{(x^2+4)'}{(x^2+4)^2}dx\)
\(\displaystyle =-\frac{1}{2(x^2+4)}+C\)
(9)\(\displaystyle \int\frac{1}{(x^2+4)^2}dx\)
\(\displaystyle =\frac{1}{4}\int\frac{(x^2+4)-x^2}{(x^2+4)^2}dx\)
\(\displaystyle =\frac{1}{4}\int\frac{1}{x^2+4}dx-\frac{1}{4}\int\frac{x^2}{(x^2+4)^2}dx\)
\(\displaystyle =\frac{1}{4}\int\frac{1}{x^2+4}dx\)
\(\displaystyle \ \ \ -\frac{1}{4}\left\{-\frac{x}{2(x^2+4)}+\frac{1}{2}\int\frac{1}{x^2+4}dx\right\}\)
\(\displaystyle =\frac{1}{8}\int\frac{1}{x^2+4}dx+\frac{x}{8(x^2+4)}\)
\(\displaystyle =\frac{1}{16}\tan^{-1}\frac{x}{2}+\frac{x}{8(x^2+4)}+C\)
\(\displaystyle =\frac{1}{4}\int\frac{1}{x^2+4}dx-\frac{1}{4}\int\frac{x^2}{(x^2+4)^2}dx\)
\(\displaystyle =\frac{1}{4}\int\frac{1}{x^2+4}dx\)
\(\displaystyle \ \ \ -\frac{1}{4}\left\{-\frac{x}{2(x^2+4)}+\frac{1}{2}\int\frac{1}{x^2+4}dx\right\}\)
\(\displaystyle =\frac{1}{8}\int\frac{1}{x^2+4}dx+\frac{x}{8(x^2+4)}\)
\(\displaystyle =\frac{1}{16}\tan^{-1}\frac{x}{2}+\frac{x}{8(x^2+4)}+C\)
(10)\(\displaystyle \int\frac{1}{(x^2+9)^3}dx\)
\(\displaystyle I_n=\int\frac{1}{(x^2+9)^n}dx\)とおくと、
\(\displaystyle I_{n+1}=\frac{1}{18n}\left\{\frac{x}{(x^2+9)^n}+(2n-1)I_n\right\}\)
\(\displaystyle =\frac{x}{18n(x^2+9)^n}+\frac{2n-1}{18n}I_n\)
この漸化式を順番に解く。
\(\displaystyle I_1=\int\frac{1}{x^2+9}dx\)
\(\displaystyle =\frac{1}{3}\tan^{-1}\frac{x}{3}+C\)
\(\displaystyle I_2=\frac{x}{18(x^2+9)}+\frac{1}{18}I_1\)
\(\displaystyle =\frac{x}{18(x^2+9)}+\frac{1}{54}\tan^{-1}\frac{x}{3}+C\)
\(\displaystyle I_3=\frac{x}{36(x^2+9)^2}+\frac{1}{12}I_2\)
\(\displaystyle =\frac{x}{36(x^2+9)^2}+\frac{x}{216(x^2+9)}\)
\(\displaystyle \ \ \ +\frac{1}{648}\tan^{-1}\frac{x}{3}+C\)
よって、
\(\displaystyle \int\frac{1}{(x^2+9)^3}dx=\frac{x}{36(x^2+9)^2}\)
\(\displaystyle \ \ \ +\frac{x}{216(x^2+9)}+\frac{1}{648}\tan^{-1}\frac{x}{3}+C\)
\(\displaystyle I_{n+1}=\frac{1}{18n}\left\{\frac{x}{(x^2+9)^n}+(2n-1)I_n\right\}\)
\(\displaystyle =\frac{x}{18n(x^2+9)^n}+\frac{2n-1}{18n}I_n\)
この漸化式を順番に解く。
\(\displaystyle I_1=\int\frac{1}{x^2+9}dx\)
\(\displaystyle =\frac{1}{3}\tan^{-1}\frac{x}{3}+C\)
\(\displaystyle I_2=\frac{x}{18(x^2+9)}+\frac{1}{18}I_1\)
\(\displaystyle =\frac{x}{18(x^2+9)}+\frac{1}{54}\tan^{-1}\frac{x}{3}+C\)
\(\displaystyle I_3=\frac{x}{36(x^2+9)^2}+\frac{1}{12}I_2\)
\(\displaystyle =\frac{x}{36(x^2+9)^2}+\frac{x}{216(x^2+9)}\)
\(\displaystyle \ \ \ +\frac{1}{648}\tan^{-1}\frac{x}{3}+C\)
よって、
\(\displaystyle \int\frac{1}{(x^2+9)^3}dx=\frac{x}{36(x^2+9)^2}\)
\(\displaystyle \ \ \ +\frac{x}{216(x^2+9)}+\frac{1}{648}\tan^{-1}\frac{x}{3}+C\)
次の学習に進もう！