Question

(-x平方-2)/(x2+x+1)2的不定积分,2表示平方

Answer 1

∫[(-x²-2)/(x²+x+1)²]dx

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