次の3つの積分を計算します。 (1) $\int_{1/2}^{1/\sqrt{2}} \frac{1}{\sqrt{1-x^2}} dx$ (2) $\int (x^2+x+1) \log x dx$ (3) $\int \frac{1}{(x+2)(x^3+8)} dx$

1. 問題の内容

次の3つの積分を計算します。

(1) $\int_{1/2}^{1/\sqrt{2}} \frac{1}{\sqrt{1-x^2}} dx$

(2) $\int (x^2+x+1) \log x dx$

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

2. 解き方の手順

(1)

$\int \frac{1}{\sqrt{1-x^2}} dx = \arcsin x + C$

したがって、

$\int_{1/2}^{1/\sqrt{2}} \frac{1}{\sqrt{1-x^2}} dx = \arcsin (1/\sqrt{2}) - \arcsin (1/2) = \pi/4 - \pi/6 = \pi/12$

(2)

部分積分を用いて計算します。

$\int (x^2+x+1) \log x dx = \int (x^2+x+1) \log x dx$

$u = \log x$, $dv = (x^2+x+1)dx$ とおくと、

$du = \frac{1}{x} dx$, $v = \frac{x^3}{3} + \frac{x^2}{2} + x$

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

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

$= (\frac{x^3}{3} + \frac{x^2}{2} + x) \log x - (\frac{x^3}{9} + \frac{x^2}{4} + x) + C$

(3)

$\frac{1}{(x+2)(x^3+8)} = \frac{1}{(x+2)(x+2)(x^2-2x+4)} = \frac{1}{(x+2)^2 (x^2-2x+4)}$

部分分数分解します。

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

$1 = A(x+2)(x^2-2x+4) + B(x^2-2x+4) + (Cx+D)(x+2)^2$

$1 = A(x^3+8) + B(x^2-2x+4) + (Cx+D)(x^2+4x+4)$

$1 = Ax^3 + 8A + Bx^2 - 2Bx + 4B + Cx^3 + 4Cx^2 + 4Cx + Dx^2 + 4Dx + 4D$

$1 = (A+C)x^3 + (B+4C+D)x^2 + (-2B+4C+4D)x + (8A+4B+4D)$

$A+C = 0$

$B+4C+D = 0$

$-2B+4C+4D = 0$

$8A+4B+4D = 1$

$C = -A$

$B-4A+D = 0$

$-2B-4A+4D = 0$

$8A+4B+4D = 1$

$B+D = 4A$

$2B-4D = -4A$

$B-2D = -2A$

$D = 4A-B$

$B-2(4A-B) = -2A$

$B-8A+2B = -2A$

$3B = 6A$

$B = 2A$

$D = 4A-2A = 2A$

$8A+4(2A)+4(2A) = 1$

$8A+8A+8A = 1$

$24A = 1$

$A = \frac{1}{24}$

$B = \frac{2}{24} = \frac{1}{12}$

$C = -\frac{1}{24}$

$D = \frac{2}{24} = \frac{1}{12}$

$\int \frac{1}{(x+2)^2 (x^2-2x+4)} dx = \int (\frac{1/24}{x+2} + \frac{1/12}{(x+2)^2} + \frac{-1/24 x+1/12}{x^2-2x+4}) dx$

$= \frac{1}{24} \log |x+2| - \frac{1}{12(x+2)} + \int \frac{-1/24 x+1/12}{x^2-2x+4} dx$

$= \frac{1}{24} \log |x+2| - \frac{1}{12(x+2)} + \int \frac{-1/24 (x-1) + 1/12 - 1/24}{x^2-2x+4} dx$

$= \frac{1}{24} \log |x+2| - \frac{1}{12(x+2)} - \frac{1}{48} \log (x^2-2x+4) + \frac{1}{24}\int \frac{1}{(x-1)^2+3}dx$

$= \frac{1}{24} \log |x+2| - \frac{1}{12(x+2)} - \frac{1}{48} \log (x^2-2x+4) + \frac{1}{24\sqrt{3}}\arctan(\frac{x-1}{\sqrt{3}}) + C$

3. 最終的な答え

(1) $\frac{\pi}{12}$

(2) $(\frac{x^3}{3} + \frac{x^2}{2} + x) \log x - (\frac{x^3}{9} + \frac{x^2}{4} + x) + C$

(3) $\frac{1}{24} \log |x+2| - \frac{1}{12(x+2)} - \frac{1}{48} \log (x^2-2x+4) + \frac{1}{24\sqrt{3}}\arctan(\frac{x-1}{\sqrt{3}}) + C$