部分分式積分法,即通過將原函數拆分為部分分式來簡化積分步驟,是計算積分時的一個常用技巧。任何有理函數都可拆分為多個多項式和部分分式的和,每個部分分式中的分子次數小於分母,然後根據積分表及利用其他積分技巧,將每個部分分式積分,就得到原函數的積分。 以下是一個簡單的例子。計算 ∫ 10 x 2 + 12 x + 20 x 3 − 8 d x {\displaystyle \int {10x^{2}+12x+20 \over x^{3}-8}\,dx} 時,需要先將它拆分為部分分式: 10 x 2 + 12 x + 20 x 3 − 8 = 10 x 2 + 12 x + 20 ( x − 2 ) ( x 2 + 2 x + 4 ) = A x − 2 + B x + C x 2 + 2 x + 4 {\displaystyle {10x^{2}+12x+20 \over x^{3}-8}={10x^{2}+12x+20 \over (x-2)(x^{2}+2x+4)}={A \over x-2}+{Bx+C \over x^{2}+2x+4}} 通分得到: 10 x 2 + 12 x + 20 = A ( x 2 + 2 x + 4 ) + ( B x + C ) ( x − 2 ) {\displaystyle 10x^{2}+12x+20=A(x^{2}+2x+4)+(Bx+C)(x-2)\,} 整理,原式變為: 10 x 2 + 12 x + 20 = ( A + B ) x 2 + ( 2 A − 2 B + C ) x + ( 4 A − 2 C ) {\displaystyle 10x^{2}+12x+20=(A+B)x^{2}+(2A-2B+C)x+(4A-2C)\,} 因此, A + B = 10 {\displaystyle A+B=10\,} 2 A − 2 B + C = 12 {\displaystyle 2A-2B+C=12\,} 4 A − 2 C = 20 {\displaystyle 4A-2C=20\,} 解方程組,得到: A = 7 {\displaystyle A=7\,} B = 3 {\displaystyle B=3\,} C = 4 {\displaystyle C=4\,} 所以: 10 x 2 + 12 x + 20 x 3 − 8 = 7 x − 2 + 3 x + 4 x 2 + 2 x + 4 {\displaystyle {10x^{2}+12x+20 \over x^{3}-8}={7 \over x-2}+{3x+4 \over x^{2}+2x+4}} 即: ∫ 10 x 2 + 12 x + 20 x 3 − 8 d x = ∫ ( 7 x − 2 + 3 x + 4 x 2 + 2 x + 4 ) d x = ∫ 7 x − 2 d x + ∫ 3 x + 4 x 2 + 2 x + 4 d x {\displaystyle \int {10x^{2}+12x+20 \over x^{3}-8}\,dx=\int ({7 \over x-2}+{3x+4 \over x^{2}+2x+4})\,dx=\int {7 \over x-2}\,dx+\int {3x+4 \over x^{2}+2x+4}\,dx} 利用換元積分法,將 x − 2 {\displaystyle x-2\,} 與 x 2 + 2 x + 4 {\displaystyle x^{2}+2x+4\,} 分別換元,便得到結果: ∫ 10 x 2 + 12 x + 20 x 3 − 8 d x {\displaystyle \int {10x^{2}+12x+20 \over x^{3}-8}\,dx} = 7 ln | x − 2 | + ∫ 3 2 ( 2 x + 2 ) + 1 x 2 + 2 x + 4 d x {\displaystyle =7\ln |x-2|+\int {{{\frac {3}{2}}(2x+2)+1} \over x^{2}+2x+4}\,dx} = 7 ln | x − 2 | + 3 2 ∫ 2 x + 2 x 2 + 2 x + 4 d x + ∫ 1 ( x + 1 ) 2 + 3 d x {\displaystyle =7\ln |x-2|+{\frac {3}{2}}\int {2x+2 \over x^{2}+2x+4}\,dx+\int {1 \over (x+1)^{2}+3}\,dx} = 7 ln | x − 2 | + 3 2 ln | x 2 + 2 x + 4 | + 1 3 arctan ( x + 1 3 ) + C {\displaystyle =7\ln |x-2|+{\frac {3}{2}}\ln |x^{2}+2x+4|+{\frac {1}{\sqrt {3}}}\arctan({x+1 \over {\sqrt {3}}})+C} 拆分為部分分式 (頁面存檔備份,存於互聯網檔案館) 在線積分器 (頁面存檔備份,存於互聯網檔案館) Wikiwand in your browser!Seamless Wikipedia browsing. On steroids.Every time you click a link to Wikipedia, Wiktionary or Wikiquote in your browser's search results, it will show the modern Wikiwand interface.Wikiwand extension is a five stars, simple, with minimum permission required to keep your browsing private, safe and transparent.Wikiwand for ChromeWikiwand for EdgeWikiwand for Firefox
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