$$$\frac{\sqrt{x}}{\sqrt{a^{2} - u^{2}}}$$$ の $$$x$$$ に関する積分

$$$\int \frac{\sqrt{x}}{\sqrt{a^{2} - u^{2}}}\, dx$$$ を求めよ。

定数倍の法則 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ を、$$$c=\frac{1}{\sqrt{a^{2} - u^{2}}}$$$ と $$$f{\left(x \right)} = \sqrt{x}$$$ に対して適用する:

$${\color{red}{\int{\frac{\sqrt{x}}{\sqrt{a^{2} - u^{2}}} d x}}} = {\color{red}{\frac{\int{\sqrt{x} d x}}{\sqrt{a^{2} - u^{2}}}}}$$

$$$n=\frac{1}{2}$$$ を用いて、べき乗の法則 $$$\int x^{n}\, dx = \frac{x^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ を適用します:

$$\frac{{\color{red}{\int{\sqrt{x} d x}}}}{\sqrt{a^{2} - u^{2}}}=\frac{{\color{red}{\int{x^{\frac{1}{2}} d x}}}}{\sqrt{a^{2} - u^{2}}}=\frac{{\color{red}{\frac{x^{\frac{1}{2} + 1}}{\frac{1}{2} + 1}}}}{\sqrt{a^{2} - u^{2}}}=\frac{{\color{red}{\left(\frac{2 x^{\frac{3}{2}}}{3}\right)}}}{\sqrt{a^{2} - u^{2}}}$$

したがって、

$$\int{\frac{\sqrt{x}}{\sqrt{a^{2} - u^{2}}} d x} = \frac{2 x^{\frac{3}{2}}}{3 \sqrt{a^{2} - u^{2}}}$$

積分定数を加える:

$$\int{\frac{\sqrt{x}}{\sqrt{a^{2} - u^{2}}} d x} = \frac{2 x^{\frac{3}{2}}}{3 \sqrt{a^{2} - u^{2}}}+C$$

$$$\int \frac{\sqrt{x}}{\sqrt{a^{2} - u^{2}}}\, dx = \frac{2 x^{\frac{3}{2}}}{3 \sqrt{a^{2} - u^{2}}} + C$$$A