1⼀x^2+x+1的不定积分

= ∫ d(x^2) / 2 / (1+ x^2)^1/2
= (1+ x^2)^1/2 + C

∫dx/(x²+x+1)
=4∫dx/(4x²+4x+1+3)
=4∫dx/[(2x+1)²+3]

= 4/3∫dx/{[(2x+1)/√3]²+1}
= 2/√3∫d[(2x+1)/√3]/{[(2x+1)/√3]²+1}
=2arctan[(2x+1)/√3]/√3+C

∫dx/[(1+x)(1+x²)]=(1/2)∫dx/(1+x)+(1/2)∫(1-x)/(1+x²)dx=(1/2)∫dx/(1+x)+(1/2)∫dx/(1+x²)-(1/2)∫x/(1+x²)dx=(1/2)∫d(1+x)/(1+x)+(1/2)∫dx/(1+x²)-(1/4)∫d(1+x²)/(1+x²)=(1/2)ln(1+x)+(1/2)arctanx-(1/4)ln(1+x²)+C=(1/2)arctanx+ln[√(1+x)/(1+x²)^¼]+C分式分裂过程看下面：_________________________________________________令1/[(1+x)(1+x²)]=A/(1+x)+(Bx+C)/(1+x²)1=A(1+x²)+(Bx+C)(1+x)1=A+Ax²+Bx+Bx²+Cx+C1=(A+B)x²+(B+C)x+(A+C){A+B=0→B=-A{B+C=0→C=-B=A{A+C=1→C=1-AC=1-A→A=1-A→2A=1→A=1/2=C，B=-1/2