计算1/(x^4+1)的原函数 求不定积分\int\frac{1}{x^4+1}dx. 解: 因为 \begin{aligned} \frac{1}{x^4+1}=&\frac{1}{(x^4+2x^2+1)-2x^2}=\frac{1}{(x^2+1)^2-2x^2}\\ =&\frac{1}{(x^2+\sqrt{2}x+1)(x^2-\sqr… GaryG...发表于写给学生的...打开...