Question

求1/(x^4+x^2+1)的原函数，要求要有过程

Answer 1

设t=x^2，得2t/t^2+t+1dt=2t+1-1/手仔t^2+t+1dt=2t+1/t^2+t+1dt-1/枣薯帆t^2+t+1dt接下去你自己做吧，结凳雹果等于ln（x^4+x^2+1)-2/根号3*arctan(2/根号3*x^2+1/根号3)

Answer 2

∫1/(x⁴+x²+1) dx
= (1/2)∫(x+1)/(x²+x+1) dx - (1/2)∫(x-1)/(x²-x+1) dx
= (1/4)[∫(2x+1)/(x²+x+1) dx + ∫dx/(x²+x+1)] - (1/4)[∫(2x-1)/(x²-x+1) dx - ∫dx/(x²-x+1)]
= (1/携铅4)ln|x²+x+1| - (1/4)ln|x²-x+1| + (1/4)∫dx/[(x+1/2)²+3/4] + (1/4)∫dx/[(x-1/2)²+3/4]
= (1/4)ln| (x²+x+1)/(x²-x+1) | + (1/4)(2/√3){arctan[(2x+1)√3] + arctan[(2x-1)/√3] + C
= (1/4)ln| (x²+x+1)/(x²辩态好-x+1) | + 1/(2√3)*arctan[√3x/(1-x²闭拆)] + C

公式：arctanx+arctany = arctan[(x+y)/(1-xy)]，xy<1

部分分式法拆解步骤：
x⁴+x²+1 = (x²+x+1)(x²-x+1)
令1/(x⁴+x²+1) = (Ax+B)/(x²+x+1) + (Cx+D)/(x²-x+1)
1 = (Ax+B)(x²-x+1) + (Cx+D)(x²+x+1)
1 = (A+C)x³ + (C+D-A+B)x² + (C+D+A-B)x + (B+D)
A+C=0，B+D=1
1-A-A=0 => A=1/2 C=-1/2
-1/2+1/2-B+1-B=0 => B=1/2 => D=1/2
1/(x⁴+x²+1) = (x+1)/[2(x²+x+1)] - (x-1)/[2(x²-x+1)]