## 問題１：不定積分を計算せよ

## 問題１：不定積分を計算せよ

与えられた5つの不定積分を計算します。

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

(2) $\int \frac{x^3+x^2}{(x^2+4)^2} dx$

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

(4) $\int \frac{1}{\sin x - \cos x + 1} dx$

(5) $\int \frac{\sin x}{\sin x + 1} dx$

## 解き方の手順

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

部分分数分解を行います。

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

両辺に $x(x-1)^3$ をかけると、

$x^3 + 1 = A(x-1)^3 + Bx(x-1)^2 + Cx(x-1) + Dx$

$x=0$ のとき、$1 = A(-1)^3 \implies A=-1$

$x=1$ のとき、$2 = D \implies D=2$

$x=2$ のとき、$9 = A(1)^3 + B(2)(1)^2 + C(2)(1) + D(2) = -1 + 2B + 2C + 4 \implies 2B + 2C = 6 \implies B+C = 3$

$x=-1$ のとき、$0 = A(-2)^3 + B(-1)(-2)^2 + C(-1)(-2) + D(-1) = 8 - 4B + 2C - 2 \implies -4B+2C = -6 \implies -2B + C = -3$

$B+C = 3$ と $-2B+C = -3$ より、引くと $3B = 6 \implies B = 2$.

したがって、$C = 3-B = 3-2 = 1$.

よって、

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

したがって、

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

$= -\ln|x| + 2\ln|x-1| - \frac{1}{x-1} - \frac{1}{(x-1)^2} + C$

### (2) $\int \frac{x^3+x^2}{(x^2+4)^2} dx$

$\int \frac{x^3+x^2}{(x^2+4)^2} dx = \int \frac{x(x^2+4) + x^2 - 4x}{(x^2+4)^2} dx = \int \frac{x}{x^2+4} dx + \int \frac{x^2-4x}{(x^2+4)^2} dx$

$I_1 = \int \frac{x}{x^2+4} dx = \frac{1}{2} \ln(x^2+4) + C_1$

$I_2 = \int \frac{x^2-4x}{(x^2+4)^2} dx = \int \frac{x^2+4 -4 -4x}{(x^2+4)^2} dx = \int \frac{1}{x^2+4} dx - \int \frac{4+4x}{(x^2+4)^2} dx = \frac{1}{2} \arctan(\frac{x}{2}) - \int \frac{4}{(x^2+4)^2} dx - \int \frac{4x}{(x^2+4)^2} dx = \frac{1}{2} \arctan(\frac{x}{2}) - \int \frac{4}{(x^2+4)^2} dx + \frac{2}{x^2+4}$

$\int \frac{1}{(x^2+a^2)^2} dx = \frac{x}{2a^2(x^2+a^2)} + \frac{1}{2a^3} \arctan(\frac{x}{a}) + C$ を使うと、

$\int \frac{4}{(x^2+4)^2} dx = 4 \int \frac{1}{(x^2+2^2)^2} dx = 4(\frac{x}{2(4)(x^2+4)} + \frac{1}{2(8)} \arctan(\frac{x}{2})) = \frac{x}{2(x^2+4)} + \frac{1}{4} \arctan(\frac{x}{2})$

よって、

$I_2 = \frac{1}{2} \arctan(\frac{x}{2}) - \frac{x}{2(x^2+4)} - \frac{1}{4} \arctan(\frac{x}{2}) + \frac{2}{x^2+4} = \frac{1}{4} \arctan(\frac{x}{2}) - \frac{x}{2(x^2+4)} + \frac{2}{x^2+4}$

$\int \frac{x^3+x^2}{(x^2+4)^2} dx = I_1 + I_2 = \frac{1}{2} \ln(x^2+4) + \frac{1}{4} \arctan(\frac{x}{2}) - \frac{x}{2(x^2+4)} + \frac{2}{x^2+4} + C = \frac{1}{2} \ln(x^2+4) + \frac{1}{4} \arctan(\frac{x}{2}) + \frac{4-x}{2(x^2+4)} + C$

### (3) $\int \frac{1}{\cos x} dx$

$\int \frac{1}{\cos x} dx = \int \frac{\cos x}{\cos^2 x} dx = \int \frac{\cos x}{1-\sin^2 x} dx$

$u=\sin x$ とおくと、$du=\cos x dx$

$\int \frac{1}{1-u^2} du = \frac{1}{2} \int (\frac{1}{1-u} + \frac{1}{1+u}) du = \frac{1}{2} (-\ln|1-u| + \ln|1+u|) + C = \frac{1}{2} \ln|\frac{1+u}{1-u}| + C = \frac{1}{2} \ln|\frac{1+\sin x}{1-\sin x}| + C$

$= \frac{1}{2} \ln|\frac{(1+\sin x)^2}{1-\sin^2 x}| + C = \frac{1}{2} \ln|\frac{(1+\sin x)^2}{\cos^2 x}| + C = \ln|\frac{1+\sin x}{\cos x}| + C = \ln|\sec x + \tan x| + C$

### (4) $\int \frac{1}{\sin x - \cos x + 1} dx$

$t = \tan(\frac{x}{2})$ とおくと、

$\sin x = \frac{2t}{1+t^2}, \cos x = \frac{1-t^2}{1+t^2}, dx = \frac{2}{1+t^2} dt$

$\int \frac{1}{\sin x - \cos x + 1} dx = \int \frac{1}{\frac{2t}{1+t^2} - \frac{1-t^2}{1+t^2} + 1} \frac{2}{1+t^2} dt = \int \frac{2}{2t - (1-t^2) + (1+t^2)} dt = \int \frac{2}{2t - 1 + t^2 + 1 + t^2} dt = \int \frac{2}{2t^2+2t} dt = \int \frac{1}{t^2+t} dt = \int \frac{1}{t(t+1)} dt = \int (\frac{1}{t} - \frac{1}{t+1}) dt = \ln|t| - \ln|t+1| + C = \ln|\frac{t}{t+1}| + C = \ln|\frac{\tan(\frac{x}{2})}{\tan(\frac{x}{2})+1}| + C$

### (5) $\int \frac{\sin x}{\sin x + 1} dx$

$\int \frac{\sin x}{\sin x + 1} dx = \int \frac{\sin x+1 - 1}{\sin x + 1} dx = \int (1 - \frac{1}{\sin x+1}) dx = x - \int \frac{1}{\sin x+1} dx$

$\int \frac{1}{\sin x + 1} dx = \int \frac{1-\sin x}{1 - \sin^2 x} dx = \int \frac{1-\sin x}{\cos^2 x} dx = \int (\sec^2 x - \sec x \tan x) dx = \tan x - \sec x + C$

$\int \frac{\sin x}{\sin x + 1} dx = x - (\tan x - \sec x) + C = x - \tan x + \sec x + C$

## 最終的な答え

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

(2) $\frac{1}{2} \ln(x^2+4) + \frac{1}{4} \arctan(\frac{x}{2}) + \frac{4-x}{2(x^2+4)} + C$

(3) $\ln|\sec x + \tan x| + C$

(4) $\ln|\frac{\tan(\frac{x}{2})}{\tan(\frac{x}{2})+1}| + C$

(5) $x - \tan x + \sec x + C$