与えられた6つの定積分を計算する問題です。 (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つの定積分を計算する問題です。

(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)

$\int_{-2}^2 \frac{3}{x+4} dx = 3 \int_{-2}^2 \frac{1}{x+4} dx = 3 [\ln|x+4|]_{-2}^2 = 3(\ln6 - \ln2) = 3\ln\frac{6}{2} = 3\ln3$

(2)

$\int_{0}^1 \frac{1}{3x-6} dx = \frac{1}{3}\int_{0}^1 \frac{3}{3x-6} dx = \frac{1}{3}[\ln|3x-6|]_0^1 = \frac{1}{3}(\ln|-3| - \ln|-6|) = \frac{1}{3}(\ln3 - \ln6) = \frac{1}{3}\ln\frac{3}{6} = \frac{1}{3}\ln\frac{1}{2} = -\frac{1}{3}\ln2$

(3)

$\int_{1}^2 (3x^2 + \frac{5}{x} - \frac{2}{x^3}) dx = \int_{1}^2 (3x^2 + \frac{5}{x} - 2x^{-3}) dx = [x^3 + 5\ln|x| - 2\cdot\frac{x^{-2}}{-2}]_{1}^2 = [x^3 + 5\ln|x| + \frac{1}{x^2}]_{1}^2 = (2^3 + 5\ln2 + \frac{1}{2^2}) - (1^3 + 5\ln1 + \frac{1}{1^2}) = (8 + 5\ln2 + \frac{1}{4}) - (1 + 0 + 1) = 8 + 5\ln2 + \frac{1}{4} - 2 = 6 + 5\ln2 + \frac{1}{4} = \frac{25}{4} + 5\ln2$

(4)

$\int_{1}^3 \frac{x^3-x+2}{x^2} dx = \int_{1}^3 (x - \frac{1}{x} + \frac{2}{x^2}) dx = [\frac{x^2}{2} - \ln|x| - \frac{2}{x}]_{1}^3 = (\frac{3^2}{2} - \ln3 - \frac{2}{3}) - (\frac{1^2}{2} - \ln1 - \frac{2}{1}) = (\frac{9}{2} - \ln3 - \frac{2}{3}) - (\frac{1}{2} - 0 - 2) = \frac{9}{2} - \frac{2}{3} - \ln3 - \frac{1}{2} + 2 = \frac{8}{2} - \frac{2}{3} + 2 - \ln3 = 4 - \frac{2}{3} + 2 - \ln3 = 6 - \frac{2}{3} - \ln3 = \frac{18-2}{3} - \ln3 = \frac{16}{3} - \ln3$

(5)

$\int_{-3}^{-2} \frac{x+2}{x^2+x} dx = \int_{-3}^{-2} \frac{x+2}{x(x+1)} dx = \int_{-3}^{-2} (\frac{2}{x} - \frac{1}{x+1}) dx = [2\ln|x| - \ln|x+1|]_{-3}^{-2} = (2\ln|-2| - \ln|-2+1|) - (2\ln|-3| - \ln|-3+1|) = (2\ln2 - \ln1) - (2\ln3 - \ln2) = 2\ln2 - 0 - 2\ln3 + \ln2 = 3\ln2 - 2\ln3 = \ln(2^3) - \ln(3^2) = \ln8 - \ln9 = \ln\frac{8}{9}$

(6)

$\int_{-1}^0 \frac{x+8}{x^2+x-2} dx = \int_{-1}^0 \frac{x+8}{(x+2)(x-1)} dx = \int_{-1}^0 (\frac{-2}{x+2} + \frac{3}{x-1}) dx = [-2\ln|x+2| + 3\ln|x-1|]_{-1}^0 = (-2\ln|0+2| + 3\ln|0-1|) - (-2\ln|-1+2| + 3\ln|-1-1|) = (-2\ln2 + 3\ln1) - (-2\ln1 + 3\ln2) = -2\ln2 + 0 - 0 - 3\ln2 = -5\ln2$

3. 最終的な答え

(1) $3\ln3$

(2) $-\frac{1}{3}\ln2$

(3) $\frac{25}{4} + 5\ln2$

(4) $\frac{16}{3} - \ln3$

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

(6) $-5\ln2$