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

1. 問題の内容

与えられた4つの積分を計算します。

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

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

(3) $\int \sqrt{x} dx$

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

2. 解き方の手順

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

$\frac{1}{\sqrt{x}} = x^{-1/2}$ なので、

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

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

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

(3) $\int \sqrt{x} dx$

$\sqrt{x} = x^{1/2}$ なので、

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

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

$x = \sqrt{3} \tan\theta$ と置換すると、$dx = \sqrt{3} \sec^2\theta d\theta$。

$x^2 + 3 = 3 \tan^2\theta + 3 = 3(\tan^2\theta + 1) = 3 \sec^2\theta$

$\int \frac{1}{x^2+3} dx = \int \frac{1}{3\sec^2\theta} \sqrt{3} \sec^2\theta d\theta = \frac{\sqrt{3}}{3} \int d\theta = \frac{1}{\sqrt{3}} \theta + C$

ここで、$\tan\theta = \frac{x}{\sqrt{3}}$ より、$\theta = \arctan(\frac{x}{\sqrt{3}})$

よって、$\int \frac{1}{x^2+3} dx = \frac{1}{\sqrt{3}} \arctan(\frac{x}{\sqrt{3}}) + C$

3. 最終的な答え

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

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

(3) $\frac{2}{3} x^{3/2} + C$

(4) $\frac{1}{\sqrt{3}} \arctan(\frac{x}{\sqrt{3}}) + C$