$$$\frac{1}{\sqrt{u^{2} + 1}}$$$の積分

$$$\int \frac{1}{\sqrt{u^{2} + 1}}\, du$$$ を求めよ。

$$$\frac{1}{\sqrt{u^{2} + 1}}$$$ の不定積分は $$$\int{\frac{1}{\sqrt{u^{2} + 1}} d u} = \operatorname{asinh}{\left(u \right)}$$$ です:

$${\color{red}{\int{\frac{1}{\sqrt{u^{2} + 1}} d u}}} = {\color{red}{\operatorname{asinh}{\left(u \right)}}}$$

したがって、

$$\int{\frac{1}{\sqrt{u^{2} + 1}} d u} = \operatorname{asinh}{\left(u \right)}$$

積分定数を加える:

$$\int{\frac{1}{\sqrt{u^{2} + 1}} d u} = \operatorname{asinh}{\left(u \right)}+C$$

$$$\int \frac{1}{\sqrt{u^{2} + 1}}\, du = \operatorname{asinh}{\left(u \right)} + C$$$A