積分４ - bate's blog

$\int \frac{1}{\sqrt{x^2+a}}\,dx = \log \,\|x+\sqrt{x^2+a}\|+C$
Cは積分定数
$\sqrt{x^2+a}=t-x, \hspace{5pt} x= \frac{t^2-a}{2t}, \hspace{5pt} dx = \frac{t^2+a}{2t^2}\,dt$
$\sqrt{x^2+a}=t-x$
$\hspace{39pt} = t-\frac{t^2-a}{2t}$
$\hspace{39pt} = \frac{2t^2}{2t} - \frac{t^2-a}{2t}$
$\hspace{39pt} = \frac{t^2+a}{2t}$
$\int \frac{1}{\sqrt{x^2+a}}\,dx = \int \frac{2t}{t^2+a} (\frac{t^2+a}{2t^2}\,dt)$
$\hspace{52pt} = \int \frac{1}{t} \, dt$
$\hspace{52pt} = \log \, t + C$
$\hspace{50pt}(t = x + \sqrt{x^2+a})$
$\hspace{52pt} = \log \, \|x+\sqrt{x^2+a}\|+C$