与えられた4つの定積分を計算する。 (1) $\int_{1}^{2} \frac{1}{x(x+1)} dx$ (2) $\int_{0}^{2} \frac{x-1}{x^2+5x+4} dx$ (3) $\int_{2}^{4} \frac{1}{(x+1)(x^2-1)} dx$ (4) $\int_{-1}^{0} \frac{1}{(x-1)(x^2+1)} dx$

1. 問題の内容

与えられた4つの定積分を計算する。

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

(2) $\int_{0}^{2} \frac{x-1}{x^2+5x+4} dx$

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

(4) $\int_{-1}^{0} \frac{1}{(x-1)(x^2+1)} dx$

2. 解き方の手順

(1)

被積分関数を部分分数分解する。

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

$1 = A(x+1) + Bx$

$x = 0$ のとき $A = 1$

$x = -1$ のとき $1 = -B$, $B = -1$

よって、

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

$\int_{1}^{2} \frac{1}{x(x+1)} dx = \int_{1}^{2} (\frac{1}{x} - \frac{1}{x+1}) dx = [\ln|x| - \ln|x+1|]_{1}^{2} = [\ln|\frac{x}{x+1}|]_{1}^{2}$

$= \ln(\frac{2}{3}) - \ln(\frac{1}{2}) = \ln(\frac{2}{3}) - \ln(\frac{1}{2}) = \ln(\frac{2/3}{1/2}) = \ln(\frac{4}{3})$

(2)

被積分関数の分母を因数分解する。

$x^2+5x+4 = (x+1)(x+4)$

$\frac{x-1}{x^2+5x+4} = \frac{x-1}{(x+1)(x+4)} = \frac{A}{x+1} + \frac{B}{x+4}$

$x-1 = A(x+4) + B(x+1)$

$x = -1$ のとき $-2 = 3A$, $A = -\frac{2}{3}$

$x = -4$ のとき $-5 = -3B$, $B = \frac{5}{3}$

$\int_{0}^{2} \frac{x-1}{x^2+5x+4} dx = \int_{0}^{2} (-\frac{2}{3}\frac{1}{x+1} + \frac{5}{3}\frac{1}{x+4}) dx = [-\frac{2}{3}\ln|x+1| + \frac{5}{3}\ln|x+4|]_{0}^{2}$

$= (-\frac{2}{3}\ln(3) + \frac{5}{3}\ln(6)) - (-\frac{2}{3}\ln(1) + \frac{5}{3}\ln(4)) = -\frac{2}{3}\ln(3) + \frac{5}{3}\ln(6) - \frac{5}{3}\ln(4)$

$= -\frac{2}{3}\ln(3) + \frac{5}{3}(\ln(2) + \ln(3)) - \frac{5}{3}(2\ln(2)) = -\frac{2}{3}\ln(3) + \frac{5}{3}\ln(2) + \frac{5}{3}\ln(3) - \frac{10}{3}\ln(2)$

$= \ln(3) - \frac{5}{3}\ln(2) = \ln(3) - \ln(2^{5/3}) = \ln(\frac{3}{2^{5/3}})$

(3)

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

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

$x = 1$ のとき $1 = 4A$, $A = \frac{1}{4}$

$x = -1$ のとき $1 = -2C$, $C = -\frac{1}{2}$

$x = 0$ のとき $1 = A - B - C = \frac{1}{4} - B + \frac{1}{2}$, $B = \frac{1}{4} + \frac{1}{2} - 1 = -\frac{1}{4}$

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

$= [\frac{1}{4}\ln|x-1| - \frac{1}{4}\ln|x+1| + \frac{1}{2}\frac{1}{x+1}]_{2}^{4} = [\frac{1}{4}\ln|\frac{x-1}{x+1}| + \frac{1}{2(x+1)}]_{2}^{4}$

$= (\frac{1}{4}\ln(\frac{3}{5}) + \frac{1}{10}) - (\frac{1}{4}\ln(\frac{1}{3}) + \frac{1}{6}) = \frac{1}{4}\ln(\frac{3}{5}) - \frac{1}{4}\ln(\frac{1}{3}) + \frac{1}{10} - \frac{1}{6} = \frac{1}{4}\ln(\frac{9}{5}) - \frac{1}{15}$

(4)

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

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

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

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

$1 = \frac{1}{2}x^2 + \frac{1}{2} + Bx^2 - Bx + Cx - C$

$\frac{1}{2} + B = 0$, $B = -\frac{1}{2}$

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

$\int_{-1}^{0} \frac{1}{(x-1)(x^2+1)} dx = \int_{-1}^{0} (\frac{1}{2}\frac{1}{x-1} + \frac{-\frac{1}{2}x-\frac{1}{2}}{x^2+1}) dx = \frac{1}{2} \int_{-1}^{0} (\frac{1}{x-1} - \frac{x}{x^2+1} - \frac{1}{x^2+1}) dx$

$= \frac{1}{2}[\ln|x-1| - \frac{1}{2}\ln(x^2+1) - \arctan(x)]_{-1}^{0} = \frac{1}{2}[(\ln(1) - \frac{1}{2}\ln(1) - \arctan(0)) - (\ln(2) - \frac{1}{2}\ln(2) - \arctan(-1))]$

$= \frac{1}{2}[0 - (\ln(2) - \frac{1}{2}\ln(2) + \frac{\pi}{4})] = \frac{1}{2}[-\frac{1}{2}\ln(2) - \frac{\pi}{4}] = -\frac{1}{4}\ln(2) - \frac{\pi}{8}$

3. 最終的な答え

(1) $\ln(\frac{4}{3})$

(2) $\ln(\frac{3}{2^{5/3}})$

(3) $\frac{1}{4}\ln(\frac{9}{5}) - \frac{1}{15}$

(4) $-\frac{1}{4}\ln(2) - \frac{\pi}{8}$