2019年12月23日月曜日

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. 1x3-x=1xx2-1=1xx+1x-1Ax+Bx+1+Cx-1=Ax2-1+Bx2-x+Cx2+xx3-x=A+B+Cx2+-B+Cx-Ax3-x{A+B+C=0-B+C=0-A=1A=-1B=C-1+2B=0B=12

      よって 求める積分は、

      1x3-xdx=-xdx+121x+1dx+121x-1dx=-logx+12logx+1+12logx-1=12-logx2+logx+1+logx-1=12logx2-1x2

      (積分定数の記述は省略)


    2. x3x3-7x+6=x3-7x+6+7x-6x3-7x+6=1+7x-6x3-7x+6=1+7x-6x-1x2+x-6=1+7x-6x-1x-2x+3Ax-1+Bx-2+Cx+3=Ax2+x-6+Bx2+2x-3+Cx2-3x+2x2-7x+6=A+B+Cx2+A+2B-3Cx+-6A-3B+2Cx2-7x+6{A+B+C=0A+2B-3C=7-6A-3B+2C=-64A+5B=7-8A-5B=-6-4A=1A=-14-1+5B=7B=85C=14-85=-2720

      よって、

      x3x3-7x+6dx=x-14logx-1+85logx-2-2720logx+3

    3. 1x4-1=1x2+1x+1x-1Ax+Bx2+1+Cx+1+Dx-1=Ax+Bx2-1+Cx2+1x-1+Dx2+1x+1x4-1=Ax3-Ax+Bx2-B+Cx3-x2+x-1+Dx3+x2+x+1x4-1=A+C+Dx3+B-C+Dx2+-A+C+Dx+-B-C+Dx4-1{A+C+D=0B-C+D=0-A+C+D=0-B-C+D=1A=02B=-1B=-12D=-C-12-C-C=0C=-14D=14

      よって、

      1x4-1dx=-12arctanx-14logx+1+14logx-1=-12arctanx+14logx-1x+1

    4. Ax+Bx+1+Cx+12+Dx+13=Ax+13+Bxx+12+Cxx+1+Dxxx+13=A+Bx3+3A+2B+Cx2+3A+B+C+Dx+Axx+13A=-1B=2-3+4+C=0C=-1-3+2-1+D=0D=2

      よって、

      x3-1xx+13dx=-logx+2logx+1+1x+1-1x+12

    5. 1x3+1=1x+1x2-x+1Ax+1+Bx+Cx2-x+1=Ax2-x+1+Bx+Cx+1x3+1=A+Bx2+-A+B+Cx+A+Cx3+1{A+B=0-A+B+C=0A+C=1B=-AC=1-A-A-A+1-A=0A=13B=-13C=23

      よって、

      1x3+1dx=13logx+1-13x-2x2-x+1dx=13logx+1-13·122x-4x2-x+1dx=13logx+1-162x-1-3x2-x+1dx=13logx+1-162x-1x2-x+1dx+121x2-x+1dx=13logx+1-16logx2-x+1+121x-122+322dx=13logx+1-16logx2-x+1+12·23arctan23x-12=13logx+1-16logx2-x+1+13arctan2x-13=13logx+12x2-x+1+13arctan2x-13

    6. 1x3+12dx=x3+1-x3·3x2x3+12dx=1x3+1dx-13x·3x2x3+12dx=1x3+1dx+13ddx1x3+1xdx=1x3+1dx+13·xx3+1-131x3+1dx=231x3+1dx+x3x3+1=x3x2+1+29logx+12x2-x+1+233arctan2x-13

コード

#!/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')

入出力結果(Zsh、PowerShell、Terminal、Jupyter(IPython))

% ./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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