与えられた4つの不定積分を計算します。 (1) $\int \frac{\sqrt{x}}{1+\sqrt{x}} dx$ (2) $\int \frac{1}{1+\sqrt{x^2+1}} dx$ (3) $\int \frac{1}{x^2\sqrt{1-x^2}} dx$ (4) $\int x^2\sqrt{a^2-x^2} dx$

はい、承知いたしました。画像の問題を解きます。

1. 問題の内容

与えられた4つの不定積分を計算します。

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

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

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

(4) $\int x^2\sqrt{a^2-x^2} dx$

2. 解き方の手順

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

$\sqrt{x} = t$ と置換すると、$x = t^2$ より $dx = 2t dt$.

$\int \frac{t}{1+t} 2t dt = 2\int \frac{t^2}{1+t} dt = 2\int \frac{t^2-1+1}{1+t} dt = 2\int (t-1 + \frac{1}{1+t}) dt$

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

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

$\sqrt{x^2+1} = t-x$ と置換すると、$x^2+1 = t^2 - 2tx + x^2$ より $2tx = t^2 - 1$.

$x = \frac{t^2-1}{2t}$, $\sqrt{x^2+1} = t - x = t - \frac{t^2-1}{2t} = \frac{2t^2 - t^2 + 1}{2t} = \frac{t^2+1}{2t}$.

$dx = \frac{2t(2t) - (t^2-1)2}{4t^2}dt = \frac{4t^2 - 2t^2 + 2}{4t^2}dt = \frac{2t^2+2}{4t^2}dt = \frac{t^2+1}{2t^2}dt$.

$\int \frac{1}{1+\frac{t^2+1}{2t}} \frac{t^2+1}{2t^2}dt = \int \frac{2t}{2t+t^2+1} \frac{t^2+1}{2t^2}dt = \int \frac{2t}{(t+1)^2} \frac{(t+1)(t-1)}{2(t^2-1)+2} \frac{t^2+1}{2t^2}dt = \int \frac{1}{(t+1)^2}\frac{t^2+1}{t} dt = \int \frac{t^2+1}{(t+1)^2}\frac{t^2+1}{2t^2} dt$

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

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

$x = \sin \theta$ と置換すると、$dx = \cos \theta d\theta$.

$\int \frac{1}{\sin^2 \theta \sqrt{1-\sin^2 \theta}} \cos \theta d\theta = \int \frac{1}{\sin^2 \theta \cos \theta} \cos \theta d\theta = \int \frac{1}{\sin^2 \theta} d\theta = \int \csc^2 \theta d\theta = -\cot \theta + C = -\frac{\sqrt{1-x^2}}{x} + C$.

(4) $\int x^2\sqrt{a^2-x^2} dx$

$x = a\sin \theta$ と置換すると、$dx = a\cos \theta d\theta$.

$\int (a\sin \theta)^2 \sqrt{a^2 - (a\sin \theta)^2} a\cos \theta d\theta = \int a^2\sin^2 \theta \sqrt{a^2(1-\sin^2 \theta)} a\cos \theta d\theta = \int a^2 \sin^2 \theta a\cos \theta a\cos \theta d\theta = a^4 \int \sin^2 \theta \cos^2 \theta d\theta = a^4 \int (\sin \theta \cos \theta)^2 d\theta = a^4 \int (\frac{1}{2}\sin 2\theta)^2 d\theta = \frac{a^4}{4} \int \sin^2 2\theta d\theta = \frac{a^4}{4} \int \frac{1 - \cos 4\theta}{2} d\theta = \frac{a^4}{8} \int (1 - \cos 4\theta) d\theta = \frac{a^4}{8} (\theta - \frac{1}{4}\sin 4\theta) + C = \frac{a^4}{8}(\theta - \frac{1}{4} 2\sin 2\theta \cos 2\theta) + C = \frac{a^4}{8} (\theta - \frac{1}{2} (2\sin \theta \cos \theta) (\cos^2 \theta - \sin^2 \theta)) + C = \frac{a^4}{8} (\theta - \sin \theta \cos \theta (\cos^2 \theta - \sin^2 \theta)) + C = \frac{a^4}{8} (\arcsin(\frac{x}{a}) - \frac{x}{a}\frac{\sqrt{a^2-x^2}}{a} (\frac{a^2-x^2}{a^2} - \frac{x^2}{a^2})) + C = \frac{a^4}{8} (\arcsin(\frac{x}{a}) - \frac{x\sqrt{a^2-x^2}}{a^2}(\frac{a^2-2x^2}{a^2})) + C = \frac{a^4}{8} \arcsin(\frac{x}{a}) - \frac{x\sqrt{a^2-x^2}(a^2-2x^2)}{8} + C$

3. 最終的な答え

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

(2) 解けませんでした。

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

(4) $\frac{a^4}{8} \arcsin(\frac{x}{a}) - \frac{x(a^2-2x^2)\sqrt{a^2-x^2}}{8} + C$