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Question Number 122734 by bemath last updated on 19/Nov/20

Answered by liberty last updated on 19/Nov/20

Just applying partial faction ((3x^3 −x^2 +6x−4)/((x^2 +1)(x^2 +2))) = ((Ax+B)/(x^2 +1))+ ((Cx+D)/(x^2 +2)) gives A=3 , B=−3 , C=0 and D=2 W(x)=∫ ((3x−3)/(x^2 +1)) + (2/(x^2 +2)) dx W(x)= (3/2)∫ ((2x)/(x^2 +1))dx−∫(3/(x^2 +1))dx+∫(2/(x^2 +2))dx W(x)=((3ℓn(x^2 +1))/2)−3 tan^(−1) (x)+(√2) tan^(−1) ((x/( (√2))))+ c

Justapplyingpartialfaction3x3−x2+6x−4(x2+1)(x2+2)=Ax+Bx2+1+Cx+Dx2+2givesA=3,B=−3,C=0andD=2W(x)=∫3x−3x2+1+2x2+2dxW(x)=32∫2xx2+1dx−∫3x2+1dx+∫2x2+2dxW(x)=3ℓn(x2+1)2−3tan−1(x)+2tan−1(x2)+c

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