Question

Resolve into partial fractions : 2x
$\div$
x2+1 (x−1)​

Answer 1

Answer:

Let,

(x−1)(x

2

+1)

2x+1

=

(x−1)

A

+

(x

2

+1)

Bx+C

...(1)

(x−1)(x

2

+1)

2x+1

=

(x−1)(x

2

+1)

A(x

2

+1)+(Bx+C)(x−1)

2x+1=A(x

2

+1)+B(x

2

−x)+C(x−1)....(2)

Putting x−1=0 i.e, x=1 in equation (2)

2(1)+1=A(1

2

+1)+B(1

2

−1)+C(1−1)

3=2A+B(0)+C(0)

A=

2

3

Now, on comparing with coefficients of x

2

in equation (2)

0=A+B

On putting value of A

0=

2

3

+B

B=

2

−3

On comparing with coeff. of x in equation (2)

2=−B+C

On putting value of B

2=

2

3

+C

C=

2

4−3

⇒C=

2

1

On putting values of A,B and C is equation (1)

(x−1)(x

2

+1)

2x+1

=

2

1

[

x−1

3

−

x

2

+1

3x−1

].