(1+X^2)/(1+X^4)的不定积分

Question

(1+X^2)/(1+X^4)的不定积分

Answer 1

∫ (1 + x²)/(1 + x⁴) dx,上下除以x²
= ∫ (1/x² + 1)/(1/x² + x²) dx
= ∫ d(x - 1/x)/[(1/x)² - 2(1/x)(x) + (x)² + 2],将分子积分后移进dx里,凑微分
= ∫ d(x - 1/x)/[(x - 1/x)² + (√2)²]
根据公式∫ dx/(a² + x²) = (1/a)arctan(x/a),直接飞去答案
= (1/√2)arctan[(x - 1/x)/√2] + C
= (1/√2)arctan[x/√2 - 1/(x√2)] + C