Question

求不定积分。

求∫1/（1＋（x的四次幂））dx的不定积分。

Answer 1

过程如下：（楼上的方法很巧妙，两种方法的结果是等价的）

Answer 2

∫ 1/(1 + x⁴) dx
= (1/2)∫ [(1 + x²) + (1 - x²)]/(1 + x⁴) dx
= (1/2)∫ (1 + x²)/(1 + x⁴) dx + (1/2)∫ (1 - x²)/(1 + x⁴) dx
= (1/2)∫ (1/x² + 1)/(1/x² + x²) dx + (1/2)∫ (1/x² - 1)/(1/x² + x²) dx
= (1/2)∫ d(x - 1/x)/[(x - 1/x)² + 2] - (1/2)∫ d(x + 1/x)/[(x + 1/x)² - 2]
= (1/2)(1/√2)arctan[(x - 1/x)/√2] - (1/2)(1/2√2)ln|[(x + 1/x - √2)/(x + 1/x + √2)| + C
= (1/2√2)arctan(x/√2 - 1/x√2) - (1/4√2)ln|(x² - x√2 + 1)/(x² + x√2 + 1)| + C