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Question Number 67561 by azizullah last updated on 28/Aug/19

Answered by Rasheed.Sindhi last updated on 28/Aug/19

Let ((2x+1)/((x^2 +1)(x−1)))=((Ax+B)/(x^2 +1))+(C/(x−1)) (Ax+B)(x−1)+C(x^2 +1)=2x+1 x=1⇒2C=3⇒C=3/2 Ax^2 +Bx−Ax−B+Cx^2 +C=2x+1 (A+C)x^2 +(B−A)x+C−B=0x^2 +2x+1 A+C=0⇒A=−C=−3/2 B−A=2⇒B=A+2=−3/2 +2=1/2 C−B=1⇒3/2−1/2=1 ((2x+1)/((x^2 +1)(x−1)))=((−(3/2)x+(1/2))/(x^2 +1))+((3/2)/(x−1)) =((−3x+1)/(2(x^2 +1)))+(3/(2(x−1)))

Let2x+1(x2+1)(x−1)=Ax+Bx2+1+Cx−1(Ax+B)(x−1)+C(x2+1)=2x+1x=1⇒2C=3⇒C=3/2Ax2+Bx−Ax−B+Cx2+C=2x+1(A+C)x2+(B−A)x+C−B=0x2+2x+1A+C=0⇒A=−C=−3/2B−A=2⇒B=A+2=−3/2+2=1/2C−B=1⇒3/2−1/2=12x+1(x2+1)(x−1)=−32x+12x2+1+32x−1=−3x+12(x2+1)+32(x−1)

Commented by azizullah last updated on 29/Aug/19

Sir Alot of thanks!

SirAlotofthanks!

Commented by Rasheed.Sindhi last updated on 30/Aug/19

You're welcome mr Azizullah!

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