与えられた積分問題を解きます。問題は以下の通りです。 (1) $\int \frac{1}{(x+1)(2x+1)} dx$ (2) $\int \frac{x^2}{x^2-1} dx$ (3) $\int \frac{3x+2}{x^2(x+1)} dx$ (4) $\int \frac{1}{(x+1)^2(x+2)} dx$ (5) $\int \frac{x+5}{x(x^2+4x+5)} dx$ (6) $\int \frac{1}{x^3+1} dx$ (7) $\int \frac{x^4}{(x^2+2)^2} dx$ (8) $\int \frac{1}{(x-1)(x-2)(x-3)} dx$

1. 問題の内容

与えられた積分問題を解きます。問題は以下の通りです。

(1) $\int \frac{1}{(x+1)(2x+1)} dx$

(2) $\int \frac{x^2}{x^2-1} dx$

(3) $\int \frac{3x+2}{x^2(x+1)} dx$

(4) $\int \frac{1}{(x+1)^2(x+2)} dx$

(5) $\int \frac{x+5}{x(x^2+4x+5)} dx$

(6) $\int \frac{1}{x^3+1} dx$

(7) $\int \frac{x^4}{(x^2+2)^2} dx$

(8) $\int \frac{1}{(x-1)(x-2)(x-3)} dx$

2. 解き方の手順

(1) $\int \frac{1}{(x+1)(2x+1)} dx$

部分分数分解を行います。

$\frac{1}{(x+1)(2x+1)} = \frac{A}{x+1} + \frac{B}{2x+1}$

$1 = A(2x+1) + B(x+1)$

$x = -1$ のとき $1 = A(-2+1) \Rightarrow A = -1$

$x = -\frac{1}{2}$ のとき $1 = B(-\frac{1}{2} + 1) \Rightarrow 1 = \frac{1}{2}B \Rightarrow B = 2$

$\int (\frac{-1}{x+1} + \frac{2}{2x+1}) dx = -\ln|x+1| + \ln|2x+1| + C$

(2) $\int \frac{x^2}{x^2-1} dx$

$\frac{x^2}{x^2-1} = \frac{x^2-1+1}{x^2-1} = 1 + \frac{1}{x^2-1} = 1 + \frac{1}{(x-1)(x+1)}$

部分分数分解: $\frac{1}{(x-1)(x+1)} = \frac{A}{x-1} + \frac{B}{x+1}$

$1 = A(x+1) + B(x-1)$

$x = 1$ のとき $1 = 2A \Rightarrow A = \frac{1}{2}$

$x = -1$ のとき $1 = -2B \Rightarrow B = -\frac{1}{2}$

$\int (1 + \frac{1/2}{x-1} - \frac{1/2}{x+1}) dx = x + \frac{1}{2}\ln|x-1| - \frac{1}{2}\ln|x+1| + C = x + \frac{1}{2}\ln|\frac{x-1}{x+1}| + C$

(3) $\int \frac{3x+2}{x^2(x+1)} dx$

部分分数分解: $\frac{3x+2}{x^2(x+1)} = \frac{A}{x} + \frac{B}{x^2} + \frac{C}{x+1}$

$3x+2 = Ax(x+1) + B(x+1) + Cx^2$

$3x+2 = Ax^2 + Ax + Bx + B + Cx^2$

$3x+2 = (A+C)x^2 + (A+B)x + B$

$B = 2, A+B = 3 \Rightarrow A = 1, A+C = 0 \Rightarrow C = -1$

$\int (\frac{1}{x} + \frac{2}{x^2} - \frac{1}{x+1}) dx = \ln|x| - \frac{2}{x} - \ln|x+1| + C = \ln|\frac{x}{x+1}| - \frac{2}{x} + C$

(4) $\int \frac{1}{(x+1)^2(x+2)} dx$

部分分数分解: $\frac{1}{(x+1)^2(x+2)} = \frac{A}{x+1} + \frac{B}{(x+1)^2} + \frac{C}{x+2}$

$1 = A(x+1)(x+2) + B(x+2) + C(x+1)^2$

$x = -1$ のとき $1 = B(1) \Rightarrow B = 1$

$x = -2$ のとき $1 = C(-1)^2 \Rightarrow C = 1$

$1 = A(x^2+3x+2) + (x+2) + (x^2+2x+1)$

$1 = Ax^2 + 3Ax + 2A + x + 2 + x^2 + 2x + 1$

$1 = (A+1)x^2 + (3A+3)x + (2A+3)$

$A+1 = 0 \Rightarrow A = -1$

$3A+3 = 0 \Rightarrow A = -1$

$2A+3 = 1 \Rightarrow 2A = -2 \Rightarrow A = -1$

$\int (\frac{-1}{x+1} + \frac{1}{(x+1)^2} + \frac{1}{x+2}) dx = -\ln|x+1| - \frac{1}{x+1} + \ln|x+2| + C = \ln|\frac{x+2}{x+1}| - \frac{1}{x+1} + C$

(5) $\int \frac{x+5}{x(x^2+4x+5)} dx$

部分分数分解: $\frac{x+5}{x(x^2+4x+5)} = \frac{A}{x} + \frac{Bx+C}{x^2+4x+5}$

$x+5 = A(x^2+4x+5) + (Bx+C)x$

$x+5 = Ax^2 + 4Ax + 5A + Bx^2 + Cx$

$x+5 = (A+B)x^2 + (4A+C)x + 5A$

$5A = 5 \Rightarrow A = 1$

$4A+C = 1 \Rightarrow 4+C = 1 \Rightarrow C = -3$

$A+B = 0 \Rightarrow B = -1$

$\int (\frac{1}{x} + \frac{-x-3}{x^2+4x+5}) dx = \int (\frac{1}{x} - \frac{x+3}{x^2+4x+5}) dx$

$\int \frac{x+3}{x^2+4x+5} dx = \frac{1}{2} \int \frac{2x+6}{x^2+4x+5} dx = \frac{1}{2} \int \frac{2x+4+2}{x^2+4x+5} dx = \frac{1}{2} \int \frac{2x+4}{x^2+4x+5} dx + \int \frac{1}{x^2+4x+5} dx$

$= \frac{1}{2} \ln|x^2+4x+5| + \int \frac{1}{(x+2)^2+1} dx = \frac{1}{2} \ln|x^2+4x+5| + \arctan(x+2) + C$

$\int (\frac{1}{x} - \frac{x+3}{x^2+4x+5}) dx = \ln|x| - \frac{1}{2} \ln|x^2+4x+5| - \arctan(x+2) + C$

(6) $\int \frac{1}{x^3+1} dx = \int \frac{1}{(x+1)(x^2-x+1)} dx$

部分分数分解: $\frac{1}{(x+1)(x^2-x+1)} = \frac{A}{x+1} + \frac{Bx+C}{x^2-x+1}$

$1 = A(x^2-x+1) + (Bx+C)(x+1)$

$1 = Ax^2 - Ax + A + Bx^2 + Bx + Cx + C$

$1 = (A+B)x^2 + (-A+B+C)x + (A+C)$

$A+B = 0, -A+B+C = 0, A+C = 1$

$B = -A, -A-A+C = 0, A+C = 1$

$-2A+C = 0, A+C = 1$

$3A = 1 \Rightarrow A = \frac{1}{3}$

$B = -\frac{1}{3}, C = \frac{2}{3}$

$\int (\frac{1/3}{x+1} + \frac{-x/3+2/3}{x^2-x+1}) dx = \frac{1}{3} \int \frac{1}{x+1} dx - \frac{1}{3} \int \frac{x-2}{x^2-x+1} dx$

$= \frac{1}{3} \ln|x+1| - \frac{1}{3} \int \frac{x-2}{x^2-x+1} dx$

$\int \frac{x-2}{x^2-x+1} dx = \frac{1}{2} \int \frac{2x-4}{x^2-x+1} dx = \frac{1}{2} \int \frac{2x-1-3}{x^2-x+1} dx = \frac{1}{2} \int \frac{2x-1}{x^2-x+1} dx - \frac{3}{2} \int \frac{1}{x^2-x+1} dx$

$= \frac{1}{2} \ln|x^2-x+1| - \frac{3}{2} \int \frac{1}{(x-\frac{1}{2})^2+\frac{3}{4}} dx = \frac{1}{2} \ln|x^2-x+1| - \frac{3}{2} \cdot \frac{2}{\sqrt{3}} \arctan(\frac{x-1/2}{\sqrt{3}/2}) + C$

$= \frac{1}{2} \ln|x^2-x+1| - \sqrt{3} \arctan(\frac{2x-1}{\sqrt{3}}) + C$

$\frac{1}{3} \ln|x+1| - \frac{1}{3}(\frac{1}{2} \ln|x^2-x+1| - \sqrt{3} \arctan(\frac{2x-1}{\sqrt{3}})) + C = \frac{1}{3}\ln|x+1| - \frac{1}{6}\ln|x^2-x+1| + \frac{\sqrt{3}}{3}\arctan(\frac{2x-1}{\sqrt{3}}) + C$

(7) $\int \frac{x^4}{(x^2+2)^2} dx$

$\frac{x^4}{(x^2+2)^2} = \frac{x^4+4x^2+4 -4x^2 -4 +4}{(x^2+2)^2} = 1 - \frac{4x^2+4}{(x^2+2)^2}= 1 - \frac{4(x^2+2)-4}{(x^2+2)^2}=1-\frac{4}{x^2+2}+\frac{4}{(x^2+2)^2}$

$\int \frac{x^4}{(x^2+2)^2} dx = \int 1 - \frac{4}{x^2+2}+\frac{4}{(x^2+2)^2} dx = x - \frac{4}{\sqrt{2}}\arctan(\frac{x}{\sqrt{2}})+\int \frac{4}{(x^2+2)^2} dx = x - 2\sqrt{2}\arctan(\frac{x}{\sqrt{2}})+\int \frac{4}{(x^2+2)^2} dx$

(8) $\int \frac{1}{(x-1)(x-2)(x-3)} dx$

$\frac{1}{(x-1)(x-2)(x-3)} = \frac{A}{x-1}+\frac{B}{x-2}+\frac{C}{x-3}$

$1 = A(x-2)(x-3)+B(x-1)(x-3)+C(x-1)(x-2)$

$x=1 \Rightarrow 1 = A(-1)(-2) \Rightarrow A = \frac{1}{2}$

$x=2 \Rightarrow 1 = B(1)(-1) \Rightarrow B = -1$

$x=3 \Rightarrow 1 = C(2)(1) \Rightarrow C = \frac{1}{2}$

$\int \frac{1}{(x-1)(x-2)(x-3)} dx = \int \frac{1/2}{x-1}+\frac{-1}{x-2}+\frac{1/2}{x-3} dx = \frac{1}{2}ln|x-1|-ln|x-2|+\frac{1}{2}ln|x-3| +C = \frac{1}{2}ln|\frac{(x-1)(x-3)}{(x-2)^2}| + C$

3. 最終的な答え

(1) $-\ln|x+1| + \ln|2x+1| + C$

(2) $x + \frac{1}{2}\ln|\frac{x-1}{x+1}| + C$

(3) $\ln|\frac{x}{x+1}| - \frac{2}{x} + C$

(4) $\ln|\frac{x+2}{x+1}| - \frac{1}{x+1} + C$

(5) $\ln|x| - \frac{1}{2} \ln|x^2+4x+5| - \arctan(x+2) + C$

(6) $\frac{1}{3}\ln|x+1| - \frac{1}{6}\ln|x^2-x+1| + \frac{\sqrt{3}}{3}\arctan(\frac{2x-1}{\sqrt{3}}) + C$

(7) $x - 2\sqrt{2}\arctan(\frac{x}{\sqrt{2}})+\int \frac{4}{(x^2+2)^2} dx$

(8) $\frac{1}{2}ln|\frac{(x-1)(x-3)}{(x-2)^2}| + C$