t^2/(1+t^4)dt求不定积分 30
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∫ t^2/(t^4 + 1) dt
= (1/2)∫ [(t^2 + 1) + (t^2 - 1)]/(t^4 + 1) dt
= (1/2)∫ (t^2 + 1)/(t^4 + 1) dt + (1/2)∫ (t^2 - 1)/(t^4 + 1) dt
= (1/2)∫ (1 + 1/t^2)/(t^2 + 1/t^2) dt + (1/2)∫ (1 - 1/t^2)/(t^2 + 1/t^2) dt，分子分母各除以t^2
= (1/2)∫ d(t - 1/t)/[(t - 1/t)^2 + 2] + (1/2)∫ d(t + 1/t)/[(t + 1/t)^2 - 2]
= (1/2)(1/√2)arctan[(t - 1/t)/√2] + (1/2)(1/(2√2))ln| [(t + 1/t) - √2]/[t + 1/t) + √2] | + C
= (√2/4)arctan[t/√2 - 1/(√2t)] + (√2/8)ln| (t² - √2t + 1)/(t² + √2t + 1) | + C
= (1/2)∫ [(t^2 + 1) + (t^2 - 1)]/(t^4 + 1) dt
= (1/2)∫ (t^2 + 1)/(t^4 + 1) dt + (1/2)∫ (t^2 - 1)/(t^4 + 1) dt
= (1/2)∫ (1 + 1/t^2)/(t^2 + 1/t^2) dt + (1/2)∫ (1 - 1/t^2)/(t^2 + 1/t^2) dt，分子分母各除以t^2
= (1/2)∫ d(t - 1/t)/[(t - 1/t)^2 + 2] + (1/2)∫ d(t + 1/t)/[(t + 1/t)^2 - 2]
= (1/2)(1/√2)arctan[(t - 1/t)/√2] + (1/2)(1/(2√2))ln| [(t + 1/t) - √2]/[t + 1/t) + √2] | + C
= (√2/4)arctan[t/√2 - 1/(√2t)] + (√2/8)ln| (t² - √2t + 1)/(t² + √2t + 1) | + C
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