https://spherical-harmonics.hateblo.jp/entry/Kyodai/1932/Kougaku

2025.01.22記

[4] $\displaystyle\int\dfrac{dx}{1+\cos x}$ ， $\displaystyle\int\dfrac{x^2+1}{x^4-5x^2+4}\, dx$ ヲ求ム．

2025.02.05記
$\displaystyle\int\dfrac{dx}{(x-\alpha)(x-\beta)}$ $=\dfrac{1}{\beta-\alpha}\log\left|\dfrac{x-\beta}{x-\alpha}\right|+(積分定数)$ である．

[解答]
$\displaystyle\int\dfrac{dx}{1+\cos x}$ $=\displaystyle\int\dfrac{2 dx}{\cos^2\dfrac{x}{2}}$ $=\tan{x}{2}+(積分定数)$
であり，
$\displaystyle\int\dfrac{x^2+1}{x^4-5x^2+4}\, dx$ $=\dfrac{1}{3}\displaystyle\int\left(\dfrac{5}{x^2-4}-\dfrac{2}{x^2-1}\right)\, dx$ $=\dfrac{5}{12}\log\left|\dfrac{x-2}{x+2}\right|-\dfrac{1}{3}\log\left|\dfrac{x-1}{x+1}\right|+(積分定数)$
である．