大学定积分，求详细过程
2个回答
展开全部
∫ 1/(x⁴ + 1) dx
= (1/2)∫ [(x² + 1) - (x² - 1)]/(x⁴ + 1) dx
= (1/2)∫ (x² + 1)/(x⁴ + 1) dx - (1/2)∫ (x² - 1)/(x⁴ + 1) dx
= (1/2)∫ (1 + 1/x²)/(x² + 1/x²) dx - (1/2)∫ (1 - 1/x²)/(x² + 1/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
= (√2/4)arctan[(x - 1/x)/√2] - (√2/8)ln|(x² - √2x + 1)/(x² + √2x + 1)| + C
= (1/2)∫ [(x² + 1) - (x² - 1)]/(x⁴ + 1) dx
= (1/2)∫ (x² + 1)/(x⁴ + 1) dx - (1/2)∫ (x² - 1)/(x⁴ + 1) dx
= (1/2)∫ (1 + 1/x²)/(x² + 1/x²) dx - (1/2)∫ (1 - 1/x²)/(x² + 1/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
= (√2/4)arctan[(x - 1/x)/√2] - (√2/8)ln|(x² - √2x + 1)/(x² + √2x + 1)| + C
本回答由提问者推荐
已赞过
已踩过<
评论
收起
你对这个回答的评价是？
推荐律师服务：
若未解决您的问题，请您详细描述您的问题，通过百度律临进行免费专业咨询
