与えられた6つの定積分を計算します。 (1) $\int_{-2}^{2} \frac{3}{x+4} dx$ (2) $\int_{0}^{1} \frac{1}{3x-6} dx$ (3) $\int_{1}^{2} (3x^2 + \frac{5}{x} - \frac{2}{x^3}) dx$ (4) $\int_{1}^{3} \frac{x^3-x+2}{x^2} dx$ (5) $\int_{-3}^{-2} \frac{x+2}{x^2+x} dx$ (6) $\int_{-1}^{0} \frac{x+8}{x^2+x-2} dx$

1. 問題の内容

与えられた6つの定積分を計算します。

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

(2) $\int_{0}^{1} \frac{1}{3x-6} dx$

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

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

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

(6) $\int_{-1}^{0} \frac{x+8}{x^2+x-2} dx$

2. 解き方の手順

(1)

$\int_{-2}^{2} \frac{3}{x+4} dx = 3 \int_{-2}^{2} \frac{1}{x+4} dx = 3 [\ln|x+4|]_{-2}^{2} = 3 (\ln|2+4| - \ln|-2+4|) = 3 (\ln 6 - \ln 2) = 3 \ln \frac{6}{2} = 3 \ln 3$

(2)

$\int_{0}^{1} \frac{1}{3x-6} dx = \frac{1}{3} \int_{0}^{1} \frac{3}{3x-6} dx = \frac{1}{3} [\ln|3x-6|]_{0}^{1} = \frac{1}{3} (\ln|3(1)-6| - \ln|3(0)-6|) = \frac{1}{3} (\ln|-3| - \ln|-6|) = \frac{1}{3} (\ln 3 - \ln 6) = \frac{1}{3} \ln \frac{3}{6} = \frac{1}{3} \ln \frac{1}{2} = -\frac{1}{3} \ln 2$

(3)

$\int_{1}^{2} (3x^2 + \frac{5}{x} - \frac{2}{x^3}) dx = \int_{1}^{2} 3x^2 dx + \int_{1}^{2} \frac{5}{x} dx - \int_{1}^{2} \frac{2}{x^3} dx = [x^3]_{1}^{2} + [5\ln|x|]_{1}^{2} - [-\frac{1}{x^2}]_{1}^{2} = (2^3 - 1^3) + (5\ln 2 - 5\ln 1) - (-\frac{1}{2^2} - (-\frac{1}{1^2})) = (8-1) + (5\ln 2 - 0) - (-\frac{1}{4} + 1) = 7 + 5\ln 2 - \frac{3}{4} = \frac{25}{4} + 5\ln 2$

(4)

$\int_{1}^{3} \frac{x^3-x+2}{x^2} dx = \int_{1}^{3} (x - \frac{1}{x} + \frac{2}{x^2}) dx = [\frac{1}{2}x^2 - \ln|x| - \frac{2}{x}]_{1}^{3} = (\frac{1}{2}(3^2) - \ln 3 - \frac{2}{3}) - (\frac{1}{2}(1^2) - \ln 1 - \frac{2}{1}) = (\frac{9}{2} - \ln 3 - \frac{2}{3}) - (\frac{1}{2} - 0 - 2) = \frac{9}{2} - \ln 3 - \frac{2}{3} - \frac{1}{2} + 2 = \frac{8}{2} - \ln 3 - \frac{2}{3} + 2 = 4 - \ln 3 - \frac{2}{3} + 2 = 6 - \frac{2}{3} - \ln 3 = \frac{16}{3} - \ln 3$

(5)

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

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

$x = 0: 2 = A(1) + 0 \implies A = 2$

$x = -1: 1 = 0 + B(-1) \implies B = -1$

$\int_{-3}^{-2} (\frac{2}{x} - \frac{1}{x+1}) dx = [2\ln|x| - \ln|x+1|]_{-3}^{-2} = (2\ln|-2| - \ln|-2+1|) - (2\ln|-3| - \ln|-3+1|) = (2\ln 2 - \ln 1) - (2\ln 3 - \ln 2) = 2\ln 2 - 0 - 2\ln 3 + \ln 2 = 3\ln 2 - 2\ln 3 = \ln 2^3 - \ln 3^2 = \ln 8 - \ln 9 = \ln \frac{8}{9}$

(6)

$\int_{-1}^{0} \frac{x+8}{x^2+x-2} dx = \int_{-1}^{0} \frac{x+8}{(x+2)(x-1)} dx = \int_{-1}^{0} (\frac{A}{x+2} + \frac{B}{x-1}) dx$

$x+8 = A(x-1) + B(x+2)$

$x = 1: 9 = A(0) + B(3) \implies B = 3$

$x = -2: 6 = A(-3) + B(0) \implies A = -2$

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

3. 最終的な答え

(1) $3\ln 3$

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

(3) $\frac{25}{4} + 5\ln 2$

(4) $\frac{16}{3} - \ln 3$

(5) $\ln \frac{8}{9}$

(6) $-5\ln 2$