(1)
∫ (1 + x)/(1 + x⁴) dx
= ∫ 1/(1 + x⁴) dx + ∫ x/(1 + x⁴) dx
= (1/2)∫ [(x² + 1) - (x² - 1)]/(x⁴ + 1) dx + (1/2)∫ 1/(x⁴ + 1) d(x²)
= (1/2)∫ (x² + 1)/(x⁴ + 1) dx - (1/2)∫ (x² - 1)/(x⁴ + 1) dx + (1/2)∫ 1/[(x²)² + 1] d(x²)
= (1/2)∫ (1 + 1/x²)/(x² + 1/x²) dx - (1/2)∫ (1 - 1/x²)/(x² + 1/x²) dx + (1/2)arctan(x²)
= (1/2)∫ d(x - 1/x)/[(x - 1/x)² + 2] - (1/2)∫ d(x + 1/x)/[(x + 1/x)² - 2] + (1/2)arctan(x²)
= (1/2)(1/√2)arctan[(x - 1/x)/√2] - (1/2)[1/(2√2)]ln|[(x + 1/x) - √2]/[(x + 1/x) + √2]| + (1/2)arctan(x²) + C
= [1/(2√2)]arctan[x/√2 - 1/(√2x)] - [1/(4√2)]ln|(x² - √2x + 1)/(x² + √2x + 1)| + (1/2)arctan(x²) + C
(2)
∫ 1/(x³ + x⁵) dx
= ∫ 1/[x³(x² + 1)] dx
= ∫ [(x² + 1) - x²]/[x³(x² + 1)] dx
= ∫ 1/x³ dx - ∫ [(x² + 1) - x²]/[x(x² + 1)] dx
= - 1/(2x²) - ∫ 1/x dx + ∫ x/(x² + 1) dx
= - 1/(2x²) - ln|x| + (1/2)ln(x² + 1) + C追问
∫ (1 + x)/(1 + x⁴) dx
= ∫ 1/(1 + x⁴) dx + ∫ x/(1 + x⁴) dx
= (1/2)∫ [(x² + 1) - (x² - 1)]/(x⁴ + 1) dx + (1/2)∫ 1/(x⁴ + 1) d(x²)
= (1/2)∫ (x² + 1)/(x⁴ + 1) dx - (1/2)∫ (x² - 1)/(x⁴ + 1) dx + (1/2)∫ 1/[(x²)² + 1] d(x²)
= (1/2)∫ (1 + 1/x²)/(x² + 1/x²) dx - (1/2)∫ (1 - 1/x²)/(x² + 1/x²) dx + (1/2)arctan(x²)
= (1/2)∫ d(x - 1/x)/[(x - 1/x)² + 2] - (1/2)∫ d(x + 1/x)/[(x + 1/x)² - 2] + (1/2)arctan(x²)
= (1/2)(1/√2)arctan[(x - 1/x)/√2] - (1/2)[1/(2√2)]ln|[(x + 1/x) - √2]/[(x + 1/x) + √2]| + (1/2)arctan(x²) + C
= [1/(2√2)]arctan[x/√2 - 1/(√2x)] - [1/(4√2)]ln|(x² - √2x + 1)/(x² + √2x + 1)| + (1/2)arctan(x²) + C
(2)
∫ 1/(x³ + x⁵) dx
= ∫ 1/[x³(x² + 1)] dx
= ∫ [(x² + 1) - x²]/[x³(x² + 1)] dx
= ∫ 1/x³ dx - ∫ [(x² + 1) - x²]/[x(x² + 1)] dx
= - 1/(2x²) - ∫ 1/x dx + ∫ x/(x² + 1) dx
= - 1/(2x²) - ln|x| + (1/2)ln(x² + 1) + C追问
(1/2)∫ d(x - 1/x)/[(x - 1/x)² + 2] 这一步为什么不是分母提出一个2 ， (1/2)(1/√2)arctan[(x - 1/x)/√2] 看这里分母是提出了根号2. 为什么？？？
追答可以先提出一个2的
只不过用了∫ dy/(y² + a²) = (1/a)arctan(y/a)
a = √2，y = x - 1/x
明白，谢了。
追答嗯
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