Math, asked by patelkrishi75, 2 months ago

resolve x square upon in (1 - X) into (1 + x square) square into partial fraction​

Answers

Answered by kunal123478
0

Answer:

The answer is

=

1

4

x

+

1

+

1

2

(

x

1

)

2

+

1

4

x

1

Explanation:

Let's factorise the denominator

x

2

1

=

(

x

+

1

)

(

x

1

)

Therefore,

x

(

x

2

1

)

(

x

1

)

=

x

(

x

+

1

)

(

x

1

)

(

x

1

)

=

x

(

x

+

1

)

(

x

1

)

2

Let's perform the decomposition into partial fractions

x

(

x

2

1

)

(

x

1

)

=

A

x

+

1

+

B

(

x

1

)

2

+

C

x

1

=

A

(

x

1

)

2

+

B

(

x

+

1

)

+

C

(

x

1

)

(

x

+

1

)

(

x

+

1

)

(

x

1

)

2

The denominators are the same, we compare the numerators

x

=

A

(

x

1

)

2

+

B

(

x

+

1

)

+

C

(

x

1

)

(

x

+

1

)

Let

x

=

1

,

,

1

=

4

A

,

,

A

=

1

4

Let

x

=

1

,

,

1

=

2

B

,

,

B

=

1

2

Coefficients of

x

2

0

=

A

+

C

,

,

C

=

A

=

1

4

Therefore,

x

(

x

2

1

)

(

x

1

)

=

1

4

x

+

1

+

1

2

(

x

1

)

2

+

1

4

x

1

Answered by anshu5415
4

Answer:

will you say the full form of i.g.i.d

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