以下の5つの不定積分を求める問題です。$C$は積分定数とします。 (1) $\int \frac{(x+1)^3}{x^3} dx$ (2) $\int x^2 \log x dx$ (3) $\int (7x+2)e^{3x} dx$ (4) $\int \log(4x-3) dx$ (5) $\int \sin 4x \cos 2x dx$

## 問題の回答

1. 問題の内容

以下の5つの不定積分を求める問題です。$C$は積分定数とします。

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

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

(3) $\int (7x+2)e^{3x} dx$

(4) $\int \log(4x-3) dx$

(5) $\int \sin 4x \cos 2x dx$

2. 解き方の手順

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

まず、$(x+1)^3$を展開します。

$(x+1)^3 = x^3 + 3x^2 + 3x + 1$

よって、

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

積分すると、

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

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

部分積分を用います。$\log x = u, x^2 dx = dv$とすると、$u' = \frac{1}{x}, v = \frac{x^3}{3}$

$\int x^2 \log x dx = \frac{x^3}{3}\log x - \int \frac{x^3}{3} \cdot \frac{1}{x} dx = \frac{x^3}{3}\log x - \int \frac{x^2}{3} dx = \frac{x^3}{3}\log x - \frac{x^3}{9} + C$

(3) $\int (7x+2)e^{3x} dx$

部分積分を用います。$7x+2 = u, e^{3x}dx = dv$とすると、$u' = 7, v = \frac{1}{3}e^{3x}$

$\int (7x+2)e^{3x} dx = (7x+2)\frac{1}{3}e^{3x} - \int 7 \cdot \frac{1}{3}e^{3x} dx = \frac{7x+2}{3}e^{3x} - \frac{7}{3} \int e^{3x} dx = \frac{7x+2}{3}e^{3x} - \frac{7}{3} \cdot \frac{1}{3}e^{3x} + C = \frac{7x+2}{3}e^{3x} - \frac{7}{9}e^{3x} + C = (\frac{7}{3}x + \frac{2}{3} - \frac{7}{9})e^{3x} + C = (\frac{7}{3}x - \frac{1}{9})e^{3x} + C=e^{3x}(\frac{21x-1}{9})+C$

$\int (7x+2)e^{3x}dx = \frac{7x}{3} e^{3x} + \frac{2}{3} e^{3x} - \frac{7}{9} e^{3x}+ C= e^{3x}(\frac{7}{3}x + \frac{6}{9} - \frac{7}{9}) + C= e^{3x}(\frac{7}{3}x - \frac{1}{9})+ C$

$\frac{7}{3}x - \frac{1}{9} = \frac{21x-1}{9}$なので係数は

$\frac{7}{3} = \frac{21}{9}$

$\frac{21}{9}x-\frac{1}{9}e^{3x} =(\frac{6}{9}x-\frac{1}{9} )e^{3x}$

$\int (7x+2)e^{3x}dx= (\frac{7}{3}x-\frac{1}{9} )e^{3x}= (\frac{21}{9}x-\frac{1}{9} )e^{3x}= e^{3x}\frac{1}{9}(21x-1) + C$

もしくは

$\int (7x+2)e^{3x}dx= \frac{e^{3x}}{3}(7x+2)-\frac{7}{9}e^{3x}+C = e^{3x}(\frac{7x}{3}+\frac{2}{3}-\frac{7}{9}) + C= e^{3x}(\frac{21x+6-7}{9}) +C = \frac{e^{3x}}{9}(21x-1)$

(4) $\int \log(4x-3) dx$

部分積分を用います。$\log(4x-3) = u, dx = dv$とすると、$u' = \frac{4}{4x-3}, v = x$

$\int \log(4x-3) dx = x\log(4x-3) - \int x \cdot \frac{4}{4x-3} dx = x\log(4x-3) - \int \frac{4x}{4x-3} dx$

$\frac{4x}{4x-3} = \frac{4x-3+3}{4x-3} = 1 + \frac{3}{4x-3}$

$\int \frac{4x}{4x-3} dx = \int (1 + \frac{3}{4x-3}) dx = x + \frac{3}{4}\log|4x-3|$

$\int \log(4x-3) dx = x\log(4x-3) - x - \frac{3}{4}\log|4x-3| + C = (x-\frac{3}{4})\log(4x-3) - x + C = (\frac{4x-3}{4})\log(4x-3) - x + C$

(5) $\int \sin 4x \cos 2x dx$

三角関数の積和の公式を用います。

$\sin A \cos B = \frac{1}{2}(\sin(A+B) + \sin(A-B))$

$\sin 4x \cos 2x = \frac{1}{2}(\sin 6x + \sin 2x)$

$\int \sin 4x \cos 2x dx = \frac{1}{2} \int (\sin 6x + \sin 2x) dx = \frac{1}{2} (-\frac{1}{6}\cos 6x - \frac{1}{2}\cos 2x) + C = -\frac{1}{12}\cos 6x - \frac{1}{4}\cos 2x + C$

3. 最終的な答え

(1) $x + 3\log|x| - \frac{3}{x} - \frac{1}{2x^2} + C$

(2) $\frac{1}{3}x^3\log x - \frac{1}{9}x^3 + C$

(3) $(\frac{7}{3}x - \frac{1}{9})e^{3x} + C$

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

(5) $-\frac{1}{12}\cos 6x - \frac{1}{4}\cos 2x + C$

上記の答えを問題文の形式に当てはめると、

(1) 1: 3, 2: 3, 3: 2

(2) 4: 3, 5: 9

(3) 6: 21, 7: 9, 8: 9, 9: 3x

(4) 10: 4, 11: 4, 12: 3, 13: 4, 14: 3

(5) 15: 1, 16: 12, 17: 6, 18: 4, 19: 2