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问题：【微积分,求不定积分∫1/(1+x^4)dx】
答案：↓↓↓ 陈竹西的回答： 网友采纳 答： 我曾经答过一样的题. 原式 =∫(x^2+1)/[2(x^4+1)]dx-∫(x^2-1)/[2(x^4+1)]dx =1/2∫(1+1/x^2)/(x^2+1/x^2)dx-1/2∫(1-1/x^2)/(x^2+1/x^2)dx =1/2∫d(x-1/x)/[(x-1/x)^2+2]-1/2∫d(x+1/x)/[(x+1/x)^2-2] =1/4∫d(x-1/x)/[(x-1/x)^2/2+1]-1/2∫d(x+1/x)/[(x+1/x+√2)(x+1/x-√2)] =√2/4*arctan[(x-1/x)/√2]-1/4∫d(x+1/x)/(x+1/x+√2)-1/4∫d(x+1/x)/(x+1/x-√2) =√2/4*arctan[(x-1/x)/√2]-1/4*ln|x+1/x+√2|-1/4*ln|x+1/x-√2|+C | 
