请问这个定积分怎么做
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∫ dx/(1 + x^4)
= (1/2)∫ [(1 + x^2) + (1 - x^2)]/(1 + x^4) dx，乘以2除以2
= (1/2)∫ (1 + x^2)/(1 + x^4) dx + (1/2)∫ (1 - x^2)/(1 + x^4) dx
= (1/2)∫ (1/x^2 + 1)/(1/x^2 + x^2) dx + (1/2)∫ (1/x^2 - 1)/(1/x^2 + x^2) dx，分子分母除以x^2
= (1/2)∫ d(x - 1/x)/[(x - 1/x)² + 2] - (1/2)∫ d(1/x + 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/(√2x)] - 1/(4√2)ln|(x² - √2x + 1)/(x² + √2x + 1)| + C
公式1：∫ dx/(x² + a²) = (1/a)arctan(x/a)
公式2：∫ dx/(x² - a²) = 1/(2a)*ln|(x - a)/(x + a)|
= (1/2)∫ [(1 + x^2) + (1 - x^2)]/(1 + x^4) dx，乘以2除以2
= (1/2)∫ (1 + x^2)/(1 + x^4) dx + (1/2)∫ (1 - x^2)/(1 + x^4) dx
= (1/2)∫ (1/x^2 + 1)/(1/x^2 + x^2) dx + (1/2)∫ (1/x^2 - 1)/(1/x^2 + x^2) dx，分子分母除以x^2
= (1/2)∫ d(x - 1/x)/[(x - 1/x)² + 2] - (1/2)∫ d(1/x + 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/(√2x)] - 1/(4√2)ln|(x² - √2x + 1)/(x² + √2x + 1)| + C
公式1：∫ dx/(x² + a²) = (1/a)arctan(x/a)
公式2：∫ dx/(x² - a²) = 1/(2a)*ln|(x - a)/(x + a)|
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