Question

Resolve 2x into Partial Fractions (2-1)(x-3)
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Answer 1

Answer:

Let 2x−3(x−1)(x2+1)2=Ax−1+Bx+Cx2+1+Dx+E(x2+1)22x-3(x-1)(x2+1)2=Ax-1+Bx+Cx2+1+Dx+E(x2+1)2. Then, 2x−3=A(x2+1)2+(Bx+C)(x−1)(x2+1)+(Dx+E)(x−1)2x-3=A(x2+1)2+(Bx+C)(x-1)(x2+1)+(Dx+E)(x-1)…………(i)

Putting x=1 in eq (i) , we get −1=A(1+1)2⇒A=−-1=A(1+1)2⇒A=-

Comparing coefficients of like powers of x on both side of (i), we have

A+B=0,C−B=0,2A+B−C+D=0,C+E−B−D=2A+B=0,C-B=0,2A+B-C+D=0,C+E-B-D=2 and A−C−E=−3A-C-E=-3.

Putting A=−14A=-14 and solving these equations, we get

B=14=C,D=14B=14=C,D=14 and E=52∴2x−3(x−1

Answer 2

Step-by-step explanation:

Let 2x−3(x−1)(x2+1)2=Ax−1+Bx+Cx2+1+Dx+E(x2+1)2. Then, 2x−3=A(x2+1)2+(Bx+C)(x−1)(x2+1)+(Dx+E)(x−1)…………(i)

Putting x=1 in eq (i) , we get −1=A(1+1)2⇒A=−

Comparing coefficients of like powers of x on both side of (i), we have

A+B=0,C−B=0,2A+B−C+D=0,C+E−B−D=2 and A−C−E=−3.

Putting A=−14 and solving these equations, we get

B=14=C,D=14 and E=52∴2x−3(x−1)(x2+1)2=−14(x−1)+x+14(x2+1)+9x+52(x2+1)2