与えられた6つの定積分の値を計算します。 (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つの定積分の値を計算します。

(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)

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

(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)

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

(4)

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

(5)

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

部分分数分解を行う。$\frac{x+2}{x(x+1)} = \frac{A}{x} + \frac{B}{x+1}$.

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

$x = 0$ のとき、$2 = A(1) + 0 \implies A=2$.

$x = -1$ のとき、$1 = 0 + B(-1) \implies B=-1$.

よって、$\frac{x+2}{x(x+1)} = \frac{2}{x} - \frac{1}{x+1}$.

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

(6)

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

部分分数分解を行う。$\frac{x+8}{(x+2)(x-1)} = \frac{A}{x+2} + \frac{B}{x-1}$.

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

$x = 1$ のとき、$9 = 0 + B(3) \implies B=3$.

$x = -2$ のとき、$6 = A(-3) + 0 \implies A=-2$.

よって、$\frac{x+8}{(x+2)(x-1)} = \frac{-2}{x+2} + \frac{3}{x-1}$.

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

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

(6) $-5\ln(2)$