以下の6つの不定積分を計算します。 (1) $\int \frac{1}{x\sqrt{1-x}} dx$ (2) $\int \sqrt{\frac{x-1}{x+1}} dx$ (3) $\int \frac{1}{x\sqrt{x^2+x+1}} dx$ (4) $\int \frac{1}{x\sqrt{4x-3-x^2}} dx$ (5) $\int \frac{1}{\cos x} dx$ (6) $\int \frac{\sin x}{1+\sin x} dx$

はい、承知いたしました。問題文に書かれている6つの積分を解きます。

1. 問題の内容

以下の6つの不定積分を計算します。

(1) $\int \frac{1}{x\sqrt{1-x}} dx$

(2) $\int \sqrt{\frac{x-1}{x+1}} dx$

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

(4) $\int \frac{1}{x\sqrt{4x-3-x^2}} dx$

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

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

2. 解き方の手順

(1) $\int \frac{1}{x\sqrt{1-x}} dx$

$u = \sqrt{1-x}$ と置換すると、$u^2 = 1-x$ より、$x = 1-u^2$。また、$dx = -2u \, du$。

したがって、

$\int \frac{1}{x\sqrt{1-x}} dx = \int \frac{1}{(1-u^2)u} (-2u) \, du = \int \frac{-2}{1-u^2} du = \int \frac{-2}{(1-u)(1+u)} du$

部分分数分解すると、$\frac{-2}{(1-u)(1+u)} = \frac{A}{1-u} + \frac{B}{1+u}$。

$-2 = A(1+u) + B(1-u)$。$u=1$のとき $-2 = 2A \Rightarrow A=-1$。$u=-1$のとき $-2 = 2B \Rightarrow B=-1$。

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

(2) $\int \sqrt{\frac{x-1}{x+1}} dx$

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

$\int \frac{x}{\sqrt{x^2-1}} dx$ について、$u = x^2-1$ と置換すると、$du = 2x dx$ より、$\int \frac{x}{\sqrt{x^2-1}} dx = \frac{1}{2} \int \frac{du}{\sqrt{u}} = \frac{1}{2} (2\sqrt{u}) = \sqrt{x^2-1}$

$\int \frac{1}{\sqrt{x^2-1}} dx = \cosh^{-1}(x) = \ln|x + \sqrt{x^2-1}|$

したがって、$\int \sqrt{\frac{x-1}{x+1}} dx = \sqrt{x^2-1} - \ln|x + \sqrt{x^2-1}| + C$

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

$x = \frac{1}{t}$ と置換すると、$dx = -\frac{1}{t^2} dt$。

$\int \frac{1}{x\sqrt{x^2+x+1}} dx = \int \frac{1}{\frac{1}{t} \sqrt{\frac{1}{t^2} + \frac{1}{t} + 1}} \left( -\frac{1}{t^2} \right) dt = \int \frac{t}{\sqrt{\frac{1+t+t^2}{t^2}}} \left( -\frac{1}{t^2} \right) dt = \int \frac{t^2}{\sqrt{1+t+t^2}} \left( -\frac{1}{t^2} \right) dt = -\int \frac{1}{\sqrt{t^2+t+1}} dt = -\int \frac{1}{\sqrt{(t+\frac{1}{2})^2 + \frac{3}{4}}} dt$

$u = t+\frac{1}{2}$ と置換すると、$du = dt$。 $-\int \frac{1}{\sqrt{u^2 + \frac{3}{4}}} du = -\sinh^{-1} \left(\frac{u}{\sqrt{\frac{3}{4}}} \right) + C = -\sinh^{-1} \left( \frac{2u}{\sqrt{3}} \right) + C = -\sinh^{-1} \left( \frac{2(t+\frac{1}{2})}{\sqrt{3}} \right) + C = -\sinh^{-1} \left( \frac{2(\frac{1}{x}+\frac{1}{2})}{\sqrt{3}} \right) + C = -\sinh^{-1} \left( \frac{\frac{2+x}{x}}{\sqrt{3}} \right) + C = -\sinh^{-1} \left( \frac{2+x}{x\sqrt{3}} \right) + C = -\ln \left| \frac{2+x}{x\sqrt{3}} + \sqrt{\frac{(2+x)^2}{3x^2}+1} \right| + C$

(4) $\int \frac{1}{x\sqrt{4x-3-x^2}} dx = \int \frac{1}{x\sqrt{1-(x-2)^2}} dx$

$x-2 = \sin \theta$ と置換すると、$x = \sin \theta + 2$。$dx = \cos \theta \, d\theta$。

$\int \frac{1}{(\sin \theta+2)\sqrt{1-\sin^2 \theta}} \cos \theta \, d\theta = \int \frac{\cos \theta}{(\sin \theta + 2)\cos \theta} d\theta = \int \frac{1}{\sin \theta + 2} d\theta$

$t = \tan \frac{\theta}{2}$ と置換すると、$\sin \theta = \frac{2t}{1+t^2}$, $d\theta = \frac{2}{1+t^2} dt$。

$\int \frac{1}{\frac{2t}{1+t^2} + 2} \frac{2}{1+t^2} dt = \int \frac{1}{\frac{2t+2+2t^2}{1+t^2}} \frac{2}{1+t^2} dt = \int \frac{2}{2t^2+2t+2} dt = \int \frac{1}{t^2+t+1} dt = \int \frac{1}{(t+\frac{1}{2})^2+\frac{3}{4}} dt = \frac{2}{\sqrt{3}} \arctan \frac{2t+1}{\sqrt{3}} + C = \frac{2}{\sqrt{3}} \arctan \frac{2\tan(\frac{\theta}{2})+1}{\sqrt{3}} + C = \frac{2}{\sqrt{3}} \arctan \frac{2\tan(\frac{\arcsin(x-2)}{2})+1}{\sqrt{3}} + C$

(5) $\int \frac{1}{\cos x} dx = \int \sec x \, dx = \ln|\sec x + \tan x| + C$

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

3. 最終的な答え

(1) $\ln \left| \frac{1-\sqrt{1-x}}{1+\sqrt{1-x}} \right| + C$

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

(3) $-\ln \left| \frac{2+x}{x\sqrt{3}} + \sqrt{\frac{(2+x)^2}{3x^2}+1} \right| + C$

(4) $\frac{2}{\sqrt{3}} \arctan \frac{2\tan(\frac{\arcsin(x-2)}{2})+1}{\sqrt{3}} + C$

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

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