Question

b) Resolve into partial fractions : x(x 1) x 1 2 2 − +​

Answer 1

Answer:

Solution:

(x+1)(x−1)2x2−x+1 can be written as =(x+1)A+(x−1)B+(x−1)2C

(x+1)(x−1)2x2−x+1 =(x+1)(x−1)2A(x−1)2+B(x2−1)+C((x+1))

x2−x+1=(A+B)x2+(−2A+C)x+A−B+C

On comparing coefficients of x2,x and constant term we get,

A+B=1        

−2A+C=−1

A−B+C=1

On solving above relations between A,B and C we get,

A=43,B=41,C=2

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