求不定积分-x^2-2/(x^2+x+1)^2

求不定积分-x^2-2/(x^2+x+1)^2

求不定积分∫[(-x²-2)/(x²+x+1)²]dx
解：原式=-∫[(x²+2)/(x²+x+1)²]dx
(x²+2)/(x²+x+1)=A/(x²+x+1)+(Bx+C)/(x²+x+1)²=[A(x²+x+1)+Bx+C]/(x²+x+1)²
故得x²+2=Ax²+(A+B)x+A+C；这是恒等式，对应项系数相等：
∴A=1；A+B=0；A+C=2；由此解得A=1，B=-1，C=1;
故原式=-{∫[1/(x²+x+1)]dx-∫(x-1)/(x²+x+1)²]dx}
=-∫dx/[(x+1/2)²+3/4]+∫xdx/(x²+x+1)²-∫dx/(x²+x+1)²
=-∫d(x+1/2)/[(x+1/2)²+3/4]+(1/2)∫d(x²+x+1)/(x²+x+1)²-(1/2)∫dx/(x²+x+1)²-∫dx/(x²+x+1)²
=-(2/√3)arctan[2(x+1/2)/√3]-1/[2(x²+x+1)]-(3/2)∫d(x+1/2)/[(x+1/2)²+3/4]²
=-(2/√3)arctan[2(x+1/2)/√3]-1/[2(x²+x+1)]-(3/2)(2/√3)arctan[2(x+1/2)/√3]+C
=-(4/√3)arctan[(2x+1)/√3]-1/[2(x²+x+1)]+C追问(3/2)∫d(x+1/2)/[(x+1/2)²+3/4]²这个的积分怎么求呢？

如以上回答内容为低俗、色情、不良、暴力、侵权、涉及违法等信息，可以点下面链接进行举报！

点此我要举报以上问答信息