题目
求有理函数的不定积分：∫x/x2＋x＋1 dx
提问时间：2020-12-30
答案
∫x/(x^2＋x＋1) dx= (1/2)∫ dln(x^2+x+1) -(1/2)∫1/(x^2＋x＋1) dx= (1/2)ln(x^2+x+1) -(1/2)∫1/(x^2＋x＋1) dx x^2＋x＋1= (x+1/2)^2+3/4letx+1/2 = (√3/2) tanadx =(√3/2) (seca)^2 da∫1/(x^2＋x＋1) dx=∫...
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