Задача 34963 ...

top22

2019-03-26 19:09:21

∫ xdx/(x-2)(x^2+x+1)

sova

2019-03-26 22:24:22

x/(x-2)(x^2+x+1) = A/(x-2) + (Mx+N)/(x^2+x+1)

x=A*(x^2+x+1)+(Mx+N)(x-2)

x=Ax^2+Ax+A+Mx^2+Nx-2Mx-2N

A+M=0
A+N-2M=1
A-2N=0

3,5A=1
A=2/7

M=-2/7

N=1/7

=(2/7) ∫dx/(x-2) - (1/7) ∫ (2x-1)/(x^2+x+1)=

=(2/7) ∫dx/(x-2) - (1/7) ∫ (2x-1)dx/((x+0,5)^2+0,75)=[x+0,5=t;dx=dt;x=t-0,5]

=(2/7) ∫dx/(x-2) - (1/7) ∫ (2(t-0,5)-1)dt/(t^2+0,75)=

=(2/7) ∫dx/(x-2) - (1/7) ∫ 2tdt/(t^2+0,75)+(2/7) ∫ dt/(t^2+0,75)=

= [b](2/7)ln|x-2| -(1/7)ln|x^2+x+1| +(2/7)*(2/sqrt(3))arctg(x+0,5)/sqrt(3)/2 + C[/b]