与えられた8つの積分問題を解きます。それぞれの積分は以下の通りです。 (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) $\int \frac{1}{(x+1)(x+2)} dx$ (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8つの積分問題を解きます。それぞれの積分は以下の通りです。

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

(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。$\frac{1}{(x+1)(2x+1)} = \frac{A}{x+1} + \frac{B}{2x+1}$ とおくと、$1 = A(2x+1) + B(x+1)$ となります。

$x = -1$ を代入すると、 $1 = A(-1) \Rightarrow A = -1$

$x = -\frac{1}{2}$ を代入すると、 $1 = B(\frac{1}{2}) \Rightarrow B = 2$

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

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

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

$\frac{1}{(x-1)(x+1)} = \frac{A}{x-1} + \frac{B}{x+1}$ とおくと、$1 = A(x+1) + B(x-1)$

$x = 1$ を代入すると、$1 = 2A \Rightarrow A = \frac{1}{2}$

$x = -1$ を代入すると、$1 = -2B \Rightarrow B = -\frac{1}{2}$

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

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

$\frac{3x+2}{x^2(x+1)} = \frac{A}{x} + \frac{B}{x^2} + \frac{C}{x+1}$ とおくと、$3x+2 = Ax(x+1) + B(x+1) + Cx^2$

$x=0$ を代入すると、$2 = B$

$x=-1$ を代入すると、$-1 = C$

$x=1$ を代入すると、$5 = 2A + 4 + (-1) \Rightarrow 2A = 2 \Rightarrow A = 1$

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

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

$\frac{1}{(x+1)(x+2)} = \frac{A}{x+1} + \frac{B}{x+2}$ とおくと、$1 = A(x+2) + B(x+1)$

$x = -1$ を代入すると、$1 = A \Rightarrow A = 1$

$x = -2$ を代入すると、$1 = -B \Rightarrow B = -1$

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

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

$\frac{x+5}{x(x^2+4x+5)} = \frac{A}{x} + \frac{Bx+C}{x^2+4x+5}$ とおくと、$x+5 = A(x^2+4x+5) + (Bx+C)x$

$x=0$ を代入すると、$5 = 5A \Rightarrow A=1$

$x+5 = x^2+4x+5 + Bx^2+Cx \Rightarrow x+5 = (1+B)x^2+(4+C)x+5$

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

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

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

$x^2+4x+5 = (x+2)^2+1$ なので、

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

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

$\frac{1}{x^3+1} = \frac{1}{(x+1)(x^2-x+1)} = \frac{A}{x+1} + \frac{Bx+C}{x^2-x+1}$ とおくと、$1 = A(x^2-x+1) + (Bx+C)(x+1)$

$x = -1$ を代入すると、$1 = 3A \Rightarrow A = \frac{1}{3}$

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

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

$\frac{1}{3}+C = 1 \Rightarrow C = \frac{2}{3}$

$\frac{1}{3}+B = 0 \Rightarrow B = -\frac{1}{3}$

$-\frac{1}{3}+B+C = 0 \Rightarrow -\frac{1}{3}-\frac{1}{3}+\frac{2}{3} = 0$

$\int \frac{1}{x^3+1} dx = \int (\frac{1/3}{x+1} - \frac{x/3 - 2/3}{x^2-x+1}) dx = \frac{1}{3}\ln|x+1| - \frac{1}{6} \int \frac{2x-4}{x^2-x+1} dx$

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

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

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

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

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

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

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

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

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

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

$x = 1$ を代入すると、$1 = A(-1)(-2) \Rightarrow 1 = 2A \Rightarrow A = \frac{1}{2}$

$x = 2$ を代入すると、$1 = B(1)(-1) \Rightarrow 1 = -B \Rightarrow B = -1$

$x = 3$ を代入すると、$1 = C(2)(1) \Rightarrow 1 = 2C \Rightarrow C = \frac{1}{2}$

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

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

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

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

(7) $x-\frac{3}{\sqrt{2}}\arctan\frac{x}{\sqrt{2}}+\frac{x}{x^2+2} + C$

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