以下の4つの不定積分を求める問題です。 (1) $\int \frac{dx}{x\sqrt{x+1}}$ (2) $\int \frac{x\sqrt{x}}{1+\sqrt{x}}dx$ (3) $\int \frac{x^2}{\sqrt{x^2+1}}dx$ (4) $\int \frac{dx}{(x+1)^2\sqrt{1-x^2}}$

はい、承知いたしました。画像にある不定積分を求めます。

1. 問題の内容

以下の4つの不定積分を求める問題です。

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

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

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

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

2. 解き方の手順

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

$x+1 = t^2$とおくと、$dx = 2t dt$、$x = t^2 - 1$。

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

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

$\sqrt{x} = t$とおくと、$x = t^2$、$dx = 2t dt$。

$\int \frac{t^2 \cdot t}{1+t}2t dt = 2\int \frac{t^4}{1+t} dt = 2\int (t^3 - t^2 + t - 1 + \frac{1}{1+t}) dt = 2(\frac{t^4}{4} - \frac{t^3}{3} + \frac{t^2}{2} - t + \log|1+t|) + C = \frac{1}{2}x^2 - \frac{2}{3}x\sqrt{x} + x - 2\sqrt{x} + 2\log|1+\sqrt{x}| + C$.

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

$x = \sinh u$とおくと、$dx = \cosh u du$、$x^2+1 = \sinh^2 u + 1 = \cosh^2 u$。

$\int \frac{\sinh^2 u}{\cosh u} \cosh u du = \int \sinh^2 u du = \int \frac{\cosh 2u - 1}{2} du = \frac{1}{4}\sinh 2u - \frac{1}{2}u + C = \frac{1}{2}\sinh u \cosh u - \frac{1}{2}u + C = \frac{1}{2}x\sqrt{x^2+1} - \frac{1}{2}\operatorname{arcsinh}x + C = \frac{1}{2}x\sqrt{x^2+1} - \frac{1}{2}\log(x + \sqrt{x^2+1}) + C$.

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

$x = \sin\theta$とおくと、$dx = \cos\theta d\theta$。

$\int \frac{\cos\theta d\theta}{(\sin\theta + 1)^2\sqrt{1-\sin^2\theta}} = \int \frac{d\theta}{(\sin\theta + 1)^2} = \int \frac{1}{(\sin\theta + 1)^2} d\theta = \int \frac{1}{(\sin\theta + 1)^2}d\theta = \tan(\frac{\theta}{2}) - \frac{1}{\tan(\theta/2) + 1} + C$.

3. 最終的な答え

(1) $\log|\frac{\sqrt{x+1}-1}{\sqrt{x+1}+1}| + C$

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

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

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