(1) 不定積分 $\int \frac{dx}{x^3+8}$ を求める。 (2) 不定積分 $\int \frac{dx}{x^4-1}$ を求める。解析学不定積分部分分数分解積分計算2025/7/301. 問題の内容(1) 不定積分 ∫dxx3+8\int \frac{dx}{x^3+8}∫x3+8dx を求める。(2) 不定積分 ∫dxx4−1\int \frac{dx}{x^4-1}∫x4−1dx を求める。2. 解き方の手順(1) ∫dxx3+8\int \frac{dx}{x^3+8}∫x3+8dxまず、x3+8x^3 + 8x3+8 を因数分解する。x3+8=(x+2)(x2−2x+4)x^3 + 8 = (x+2)(x^2 - 2x + 4)x3+8=(x+2)(x2−2x+4)次に、1x3+8=Ax+2+Bx+Cx2−2x+4\frac{1}{x^3+8} = \frac{A}{x+2} + \frac{Bx+C}{x^2-2x+4}x3+81=x+2A+x2−2x+4Bx+C と部分分数分解する。両辺に x3+8x^3+8x3+8 を掛けると1=A(x2−2x+4)+(Bx+C)(x+2)1 = A(x^2 - 2x + 4) + (Bx+C)(x+2)1=A(x2−2x+4)+(Bx+C)(x+2)1=Ax2−2Ax+4A+Bx2+2Bx+Cx+2C1 = Ax^2 - 2Ax + 4A + Bx^2 + 2Bx + Cx + 2C1=Ax2−2Ax+4A+Bx2+2Bx+Cx+2C1=(A+B)x2+(−2A+2B+C)x+(4A+2C)1 = (A+B)x^2 + (-2A+2B+C)x + (4A+2C)1=(A+B)x2+(−2A+2B+C)x+(4A+2C)係数を比較して、A+B=0A+B = 0A+B=0−2A+2B+C=0-2A+2B+C = 0−2A+2B+C=04A+2C=14A+2C = 14A+2C=1B=−AB = -AB=−A−2A−2A+C=0 ⟹ C=4A-2A - 2A + C = 0 \implies C = 4A−2A−2A+C=0⟹C=4A4A+2(4A)=1 ⟹ 12A=1 ⟹ A=1124A + 2(4A) = 1 \implies 12A = 1 \implies A = \frac{1}{12}4A+2(4A)=1⟹12A=1⟹A=121B=−112,C=412=13B = -\frac{1}{12}, C = \frac{4}{12} = \frac{1}{3}B=−121,C=124=311x3+8=1/12x+2+(−1/12)x+1/3x2−2x+4\frac{1}{x^3+8} = \frac{1/12}{x+2} + \frac{(-1/12)x + 1/3}{x^2-2x+4}x3+81=x+21/12+x2−2x+4(−1/12)x+1/3∫dxx3+8=∫1/12x+2dx+∫(−1/12)x+1/3x2−2x+4dx\int \frac{dx}{x^3+8} = \int \frac{1/12}{x+2} dx + \int \frac{(-1/12)x + 1/3}{x^2-2x+4} dx∫x3+8dx=∫x+21/12dx+∫x2−2x+4(−1/12)x+1/3dx=112∫1x+2dx+∫(−1/12)x+1/3x2−2x+4dx= \frac{1}{12} \int \frac{1}{x+2} dx + \int \frac{(-1/12)x + 1/3}{x^2-2x+4} dx=121∫x+21dx+∫x2−2x+4(−1/12)x+1/3dx=112ln∣x+2∣+∫(−1/12)x+1/3x2−2x+4dx= \frac{1}{12} \ln |x+2| + \int \frac{(-1/12)x + 1/3}{x^2-2x+4} dx=121ln∣x+2∣+∫x2−2x+4(−1/12)x+1/3dxここで、∫(−1/12)x+1/3x2−2x+4dx\int \frac{(-1/12)x + 1/3}{x^2-2x+4} dx∫x2−2x+4(−1/12)x+1/3dx を計算する。x2−2x+4=(x−1)2+3x^2-2x+4 = (x-1)^2 + 3x2−2x+4=(x−1)2+3∫(−1/12)x+1/3x2−2x+4dx=−124∫2x−4x2−2x+4dx+112∫1x2−2x+4dx\int \frac{(-1/12)x + 1/3}{x^2-2x+4} dx = -\frac{1}{24} \int \frac{2x-4}{x^2-2x+4} dx + \frac{1}{12} \int \frac{1}{x^2-2x+4} dx∫x2−2x+4(−1/12)x+1/3dx=−241∫x2−2x+42x−4dx+121∫x2−2x+41dx=−124ln∣x2−2x+4∣+112∫1(x−1)2+3dx= -\frac{1}{24} \ln|x^2-2x+4| + \frac{1}{12} \int \frac{1}{(x-1)^2+3} dx=−241ln∣x2−2x+4∣+121∫(x−1)2+31dx=−124ln∣x2−2x+4∣+112⋅13arctanx−13+C= -\frac{1}{24} \ln|x^2-2x+4| + \frac{1}{12} \cdot \frac{1}{\sqrt{3}} \arctan{\frac{x-1}{\sqrt{3}}} + C=−241ln∣x2−2x+4∣+121⋅31arctan3x−1+C=−124ln∣x2−2x+4∣+336arctanx−13+C= -\frac{1}{24} \ln|x^2-2x+4| + \frac{\sqrt{3}}{36} \arctan{\frac{x-1}{\sqrt{3}}} + C=−241ln∣x2−2x+4∣+363arctan3x−1+Cしたがって、∫dxx3+8=112ln∣x+2∣−124ln∣x2−2x+4∣+336arctanx−13+C\int \frac{dx}{x^3+8} = \frac{1}{12} \ln |x+2| -\frac{1}{24} \ln|x^2-2x+4| + \frac{\sqrt{3}}{36} \arctan{\frac{x-1}{\sqrt{3}}} + C∫x3+8dx=121ln∣x+2∣−241ln∣x2−2x+4∣+363arctan3x−1+C(2) ∫dxx4−1\int \frac{dx}{x^4-1}∫x4−1dxx4−1=(x2−1)(x2+1)=(x−1)(x+1)(x2+1)x^4-1 = (x^2-1)(x^2+1) = (x-1)(x+1)(x^2+1)x4−1=(x2−1)(x2+1)=(x−1)(x+1)(x2+1)1x4−1=Ax−1+Bx+1+Cx+Dx2+1\frac{1}{x^4-1} = \frac{A}{x-1} + \frac{B}{x+1} + \frac{Cx+D}{x^2+1}x4−11=x−1A+x+1B+x2+1Cx+D1=A(x+1)(x2+1)+B(x−1)(x2+1)+(Cx+D)(x−1)(x+1)1 = A(x+1)(x^2+1) + B(x-1)(x^2+1) + (Cx+D)(x-1)(x+1)1=A(x+1)(x2+1)+B(x−1)(x2+1)+(Cx+D)(x−1)(x+1)1=A(x3+x2+x+1)+B(x3−x2+x−1)+(Cx+D)(x2−1)1 = A(x^3+x^2+x+1) + B(x^3-x^2+x-1) + (Cx+D)(x^2-1)1=A(x3+x2+x+1)+B(x3−x2+x−1)+(Cx+D)(x2−1)1=Ax3+Ax2+Ax+A+Bx3−Bx2+Bx−B+Cx3−Cx+Dx2−D1 = Ax^3+Ax^2+Ax+A + Bx^3-Bx^2+Bx-B + Cx^3-Cx+Dx^2-D1=Ax3+Ax2+Ax+A+Bx3−Bx2+Bx−B+Cx3−Cx+Dx2−D1=(A+B+C)x3+(A−B+D)x2+(A+B−C)x+(A−B−D)1 = (A+B+C)x^3 + (A-B+D)x^2 + (A+B-C)x + (A-B-D)1=(A+B+C)x3+(A−B+D)x2+(A+B−C)x+(A−B−D)A+B+C=0A+B+C = 0A+B+C=0A−B+D=0A-B+D = 0A−B+D=0A+B−C=0A+B-C = 0A+B−C=0A−B−D=1A-B-D = 1A−B−D=12C=0 ⟹ C=02C = 0 \implies C=02C=0⟹C=0A+B=0 ⟹ B=−AA+B = 0 \implies B = -AA+B=0⟹B=−AA+A+D=0 ⟹ D=−2AA+A+D = 0 \implies D = -2AA+A+D=0⟹D=−2AA+A+2A=1 ⟹ 4A=1 ⟹ A=14A+A+2A = 1 \implies 4A = 1 \implies A = \frac{1}{4}A+A+2A=1⟹4A=1⟹A=41B=−14,C=0,D=−12B = -\frac{1}{4}, C = 0, D = -\frac{1}{2}B=−41,C=0,D=−21∫dxx4−1=∫1/4x−1dx+∫−1/4x+1dx+∫−1/2x2+1dx\int \frac{dx}{x^4-1} = \int \frac{1/4}{x-1} dx + \int \frac{-1/4}{x+1} dx + \int \frac{-1/2}{x^2+1} dx∫x4−1dx=∫x−11/4dx+∫x+1−1/4dx+∫x2+1−1/2dx=14ln∣x−1∣−14ln∣x+1∣−12arctanx+C= \frac{1}{4} \ln|x-1| - \frac{1}{4} \ln|x+1| - \frac{1}{2} \arctan{x} + C=41ln∣x−1∣−41ln∣x+1∣−21arctanx+C=14ln∣x−1x+1∣−12arctanx+C= \frac{1}{4} \ln|\frac{x-1}{x+1}| - \frac{1}{2} \arctan{x} + C=41ln∣x+1x−1∣−21arctanx+C3. 最終的な答え(1) ∫dxx3+8=112ln∣x+2∣−124ln∣x2−2x+4∣+336arctanx−13+C\int \frac{dx}{x^3+8} = \frac{1}{12} \ln |x+2| -\frac{1}{24} \ln|x^2-2x+4| + \frac{\sqrt{3}}{36} \arctan{\frac{x-1}{\sqrt{3}}} + C∫x3+8dx=121ln∣x+2∣−241ln∣x2−2x+4∣+363arctan3x−1+C(2) ∫dxx4−1=14ln∣x−1x+1∣−12arctanx+C\int \frac{dx}{x^4-1} = \frac{1}{4} \ln|\frac{x-1}{x+1}| - \frac{1}{2} \arctan{x} + C∫x4−1dx=41ln∣x+1x−1∣−21arctanx+C