【微积分,求不定积分∫1/(1+x^4)dx】
<p>问题：【微积分,求不定积分∫1/(1+x^4)dx】<p>答案：↓↓↓<p class="nav-title mt10" style="border-top:1px solid #ccc;padding-top: 10px;">陈竹西的回答：<div class="content-b">网友采纳 答： 我曾经答过一样的题. 原式 =∫(x^2+1)/dx-∫(x^2-1)/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
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