Posted by haifeng on 2014-12-23 20:25:34 last update 2022-04-26 08:56:59 | Answers (0) | 收藏
这里 $a > 0$.
\[ \int\frac{1}{t^2-a^2}\mathrm{d}t=\frac{1}{2a}\ln\biggl|\frac{t-a}{t+a}\biggr|+C, \]
\[ \int\frac{1}{t^2+a^2}\mathrm{d}t=\frac{1}{a}\arctan\frac{t}{a}+C, \]
