数学 - Python - 解析学 - 積分の計算 - 不定積分の計算 - 有理関数の積分、部分分数分解、部分積分法

学習環境

解析入門(上) (松坂和夫数学入門シリーズ 4) (松坂和夫(著)、岩波書店)の第8章(積分の計算)、8.1(不定積分の計算)、問題1の解答を求めてみる。

1. $\begin{array}{l} \frac{1}{x^{3} - x} \\ = \frac{1}{x (x^{2} - 1)} \\ = \frac{1}{x (x + 1) (x - 1)} \\ \frac{A}{x} + \frac{B}{x + 1} + \frac{C}{x - 1} \\ = \frac{A (x^{2} - 1) + B (x^{2} - x) + C (x^{2} + x)}{x^{3} - x} \\ = \frac{(A + B + C) x^{2} + (- B + C) x - A}{x^{3} - x} \\ {\begin{cases} A + B + C = 0 \\ - B + C = 0 \\ - A = 1 \end{cases} \\ A = - 1 \\ B = C \\ - 1 + 2 B = 0 \\ B = \frac{1}{2} \end{array}$
  
  よって求める積分は、
  
  $\begin{array}{l} \int \frac{1}{x^{3} - x} dx \\ = - \int x dx + \frac{1}{2} \int \frac{1}{x + 1} dx + \frac{1}{2} \int \frac{1}{x - 1} dx \\ = - \log |x| + \frac{1}{2} \log |x + 1| + \frac{1}{2} \log |x - 1| \\ = \frac{1}{2} (- \log x^{2} + \log |x + 1| + \log |x - 1|) \\ = \frac{1}{2} \log \frac{|x^{2} - 1|}{x^{2}} \end{array}$
  
  （積分定数の記述は省略）
2. $\begin{array}{l} \frac{x^{3}}{x^{3} - 7 x + 6} \\ = \frac{x^{3} - 7 x + 6 + 7 x - 6}{x^{3} - 7 x + 6} \\ = 1 + \frac{7 x - 6}{x^{3} - 7 x + 6} \\ = 1 + \frac{7 x - 6}{(x - 1) (x^{2} + x - 6)} \\ = 1 + \frac{7 x - 6}{(x - 1) (x - 2) (x + 3)} \\ \frac{A}{x - 1} + \frac{B}{x - 2} + \frac{C}{x + 3} \\ = \frac{A (x^{2} + x - 6) + B (x^{2} + 2 x - 3) + C (x^{2} - 3 x + 2)}{x^{2} - 7 x + 6} \\ = \frac{(A + B + C) x^{2} + (A + 2 B - 3 C) x + (- 6 A - 3 B + 2 C)}{x^{2} - 7 x + 6} \\ {\begin{cases} A + B + C = 0 \\ A + 2 B - 3 C = 7 \\ - 6 A - 3 B + 2 C = - 6 \end{cases} \\ 4 A + 5 B = 7 \\ - 8 A - 5 B = - 6 \\ - 4 A = 1 \\ A = - \frac{1}{4} \\ - 1 + 5 B = 7 \\ B = \frac{8}{5} \\ C = \frac{1}{4} - \frac{8}{5} = - \frac{27}{20} \end{array}$
  
  よって、
  
  $\begin{array}{l} \int \frac{x^{3}}{x^{3} - 7 x + 6} dx \\ = x - \frac{1}{4} \log |x - 1| + \frac{8}{5} \log |x - 2| - \frac{27}{20} \log |x + 3| \end{array}$
3. $\begin{array}{l} \frac{1}{x^{4} - 1} \\ = \frac{1}{(x^{2} + 1) (x + 1) (x - 1)} \\ \frac{A x + B}{x^{2} + 1} + \frac{C}{x + 1} + \frac{D}{x - 1} \\ = \frac{(A x + B) (x^{2} - 1) + C (x^{2} + 1) (x - 1) + D (x^{2} + 1) (x + 1)}{x^{4} - 1} \\ = \frac{A x^{3} - A x + B x^{2} - B + C (x^{3} - x^{2} + x - 1) + D (x^{3} + x^{2} + x + 1)}{x^{4} - 1} \\ = \frac{(A + C + D) x^{3} + (B - C + D) x^{2} + (- A + C + D) x + (- B - C + D)}{x^{4} - 1} \\ {\begin{cases} A + C + D = 0 \\ B - C + D = 0 \\ - A + C + D = 0 \\ - B - C + D = 1 \end{cases} \\ A = 0 \\ 2 B = - 1 \\ B = - \frac{1}{2} \\ D = - C \\ - \frac{1}{2} - C - C = 0 \\ C = - \frac{1}{4} \\ D = \frac{1}{4} \end{array}$
  
  よって、
  
  $\begin{array}{l} \int \frac{1}{x^{4} - 1} dx \\ = - \frac{1}{2} \arctan x - \frac{1}{4} \log |x + 1| + \frac{1}{4} \log |x - 1| \\ = - \frac{1}{2} \arctan x + \frac{1}{4} \log |\frac{x - 1}{x + 1}| \end{array}$
4. $\begin{array}{l} \frac{A}{x} + \frac{B}{x + 1} + \frac{C}{{(x + 1)}^{2}} + \frac{D}{{(x + 1)}^{3}} \\ = \frac{A {(x + 1)}^{3} + B x {(x + 1)}^{2} + C x (x + 1) + D x}{x {(x + 1)}^{3}} \\ = \frac{(A + B) x^{3} + (3 A + 2 B + C) x^{2} + (3 A + B + C + D) x + A}{x {(x + 1)}^{3}} \\ A = - 1 \\ B = 2 \\ - 3 + 4 + C = 0 \\ C = - 1 \\ - 3 + 2 - 1 + D = 0 \\ D = 2 \end{array}$
  
  よって、
  
  $\begin{array}{l} \int \frac{x^{3} - 1}{x {(x + 1)}^{3}} dx \\ = - \log |x| + 2 \log |x + 1| + \frac{1}{x + 1} - \frac{1}{{(x + 1)}^{2}} \end{array}$
5. $\begin{array}{l} \frac{1}{x^{3} + 1} \\ = \frac{1}{(x + 1) (x^{2} - x + 1)} \\ \frac{A}{x + 1} + \frac{B x + C}{x^{2} - x + 1} \\ = \frac{A (x^{2} - x + 1) + (B x + C) (x + 1)}{x^{3} + 1} \\ = \frac{(A + B) x^{2} + (- A + B + C) x + (A + C)}{x^{3} + 1} \\ {\begin{cases} A + B = 0 \\ - A + B + C = 0 \\ A + C = 1 \end{cases} \\ B = - A \\ C = 1 - A \\ - A - A + 1 - A = 0 \\ A = \frac{1}{3} \\ B = - \frac{1}{3} \\ C = \frac{2}{3} \end{array}$
  
  よって、
  
  $\begin{array}{l} \int \frac{1}{x^{3} + 1} dx \\ = \frac{1}{3} \log |x + 1| - \frac{1}{3} \int \frac{x - 2}{x^{2} - x + 1} dx \\ = \frac{1}{3} \log |x + 1| - \frac{1}{3} \cdot \frac{1}{2} \int \frac{2 x - 4}{x^{2} - x + 1} dx \\ = \frac{1}{3} \log |x + 1| - \frac{1}{6} \int \frac{2 x - 1 - 3}{x^{2} - x + 1} dx \\ = \frac{1}{3} \log |x + 1| - \frac{1}{6} \int \frac{2 x - 1}{x^{2} - x + 1} dx + \frac{1}{2} \int \frac{1}{x^{2} - x + 1} dx \\ = \frac{1}{3} \log |x + 1| - \frac{1}{6} \log (x^{2} - x + 1) + \frac{1}{2} \int \frac{1}{{(x - \frac{1}{2})}^{2} + {(\frac{\sqrt{3}}{2})}^{2}} dx \\ = \frac{1}{3} \log |x + 1| - \frac{1}{6} \log (x^{2} - x + 1) + \frac{1}{2} \cdot \frac{2}{\sqrt{3}} \arctan (\frac{2}{\sqrt{3}} (x - \frac{1}{2})) \\ = \frac{1}{3} \log |x + 1| - \frac{1}{6} \log (x^{2} - x + 1) + \frac{1}{\sqrt{3}} \arctan \frac{2 x - 1}{\sqrt{3}} \\ = \frac{1}{3} \log \frac{{(x + 1)}^{2}}{x^{2} - x + 1} + \frac{1}{\sqrt{3}} \arctan \frac{2 x - 1}{\sqrt{3}} \end{array}$
6. $\begin{array}{l} \int \frac{1}{{(x^{3} + 1)}^{2}} dx \\ = \int \frac{(x^{3} + 1) - \frac{x}{3} \cdot 3 x^{2}}{{(x^{3} + 1)}^{2}} dx \\ = \int \frac{1}{x^{3} + 1} d x - \frac{1}{3} \int \frac{x \cdot 3 x^{2}}{{(x^{3} + 1)}^{2}} dx \\ = \int \frac{1}{x^{3} + 1} dx + \frac{1}{3} \int (\frac{d}{dx} (\frac{1}{x^{3} + 1})) x dx \\ = \int \frac{1}{x^{3} + 1} dx + \frac{1}{3} \cdot \frac{x}{x^{3} + 1} - \frac{1}{3} \int \frac{1}{x^{3} + 1} dx \\ = \frac{2}{3} \int \frac{1}{x^{3} + 1} dx + \frac{x}{3 (x^{3} + 1)} \\ = \frac{x}{3 (x^{2} + 1)} + \frac{2}{9} \log \frac{{(x + 1)}^{2}}{x^{2} - x + 1} + \frac{2}{3 \sqrt{3}} \arctan \frac{2 x - 1}{\sqrt{3}} \end{array}$

#!/usr/bin/env python3 from sympy import pprint, symbols, plot, Integral, Rational, oo print('1.') x = symbols('x') fs = [1 / (x ** 3 - x), x ** 3 / (x ** 3 - 7 * x + 6), 1 / (x ** 4 - 1), (x ** 3 - 1) / (x * (x + 1) ** 3), 1 / (x ** 3 + 1), 1 / (x ** 3 + 1) ** 2] for i, f in enumerate(fs, 1): print(f'({i})') I = Integral(f, x) for o in [I, I.doit()]: pprint(o.simplify()) print() p = plot(*fs, (x, -5, 5), ylim=(-5, 5), show=False, legend=True) colors = ['red', 'green', 'blue', 'brown', 'orange', 'purple', 'pink', 'gray', 'skyblue', 'yellow'] for o, color in zip(p, colors): o.line_color = color p.show() p.save('sample1.png')

% ./sample1.py 1. (1) ⌠ ⎮ 1 ⎮ ────── dx ⎮ 3 ⎮ x - x ⌡ ⎛ 2 ⎞ log⎝x - 1⎠ -log(x) + ─────────── 2 (2) ⌠ ⎮ 3 ⎮ x ⎮ ──────────── dx ⎮ 3 ⎮ x - 7⋅x + 6 ⌡ 8⋅log(x - 2) log(x - 1) 27⋅log(x + 3) x + ──────────── - ────────── - ───────────── 5 4 20 (3) ⌠ ⎮ 1 ⎮ ────── dx ⎮ 4 ⎮ x - 1 ⌡ log(x - 1) log(x + 1) atan(x) ────────── - ────────── - ─────── 4 4 2 (4) ⌠ ⎮ 3 ⎮ x - 1 ⎮ ────────── dx ⎮ 3 ⎮ x⋅(x + 1) ⌡ ⎛ 2 ⎞ x + (-log(x) + 2⋅log(x + 1))⋅⎝x + 2⋅x + 1⎠ ─────────────────────────────────────────── 2 x + 2⋅x + 1 (5) ⌠ ⎮ 1 ⎮ ────── dx ⎮ 3 ⎮ x + 1 ⌡ ⎛√3⋅(2⋅x - 1)⎞ ⎛ 2 ⎞ √3⋅atan⎜────────────⎟ log(x + 1) log⎝x - x + 1⎠ ⎝ 3 ⎠ ────────── - ─────────────── + ───────────────────── 3 6 3 (6) ⌠ ⎮ 1 ⎮ ───────── dx ⎮ 2 ⎮ ⎛ 3 ⎞ ⎮ ⎝x + 1⎠ ⌡ ⎛ 3 ⎞ ⎛ ⎛ 2 ⎞ ⎛√3⋅(2⋅x - 1)⎞⎞ 3⋅x + ⎝x + 1⎠⋅⎜2⋅log(x + 1) - log⎝x - x + 1⎠ + 2⋅√3⋅atan⎜────────────⎟⎟ ⎝ ⎝ 3 ⎠⎠ ───────────────────────────────────────────────────────────────────────── ⎛ 3 ⎞ 9⋅⎝x + 1⎠ %

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2019年12月23日月曜日

数学 - Python - 解析学 - 積分の計算 - 不定積分の計算 - 有理関数の積分、部分分数分解、部分積分法

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