Question

2x+3(x^2-2x-3) partial fraction​

Answer 1

Let I=

Let I=x

Let I=x2

Let I=x2+2x−3

Let I=x2+2x−32x

Let I=x2+2x−32x2

Let I=x2+2x−32x2+5x−11

Let I=x2+2x−32x2+5x−11=

Let I=x2+2x−32x2+5x−11=x

Let I=x2+2x−32x2+5x−11=x2

Let I=x2+2x−32x2+5x−11=x2+2x−3

Let I=x2+2x−32x2+5x−11=x2+2x−32(x

Let I=x2+2x−32x2+5x−11=x2+2x−32(x2

Let I=x2+2x−32x2+5x−11=x2+2x−32(x2+2x−3)+x−5

Let I=x2+2x−32x2+5x−11=x2+2x−32(x2+2x−3)+x−5=2+

Let I=x2+2x−32x2+5x−11=x2+2x−32(x2+2x−3)+x−5=2+x

Let I=x2+2x−32x2+5x−11=x2+2x−32(x2+2x−3)+x−5=2+x2

Let I=x2+2x−32x2+5x−11=x2+2x−32(x2+2x−3)+x−5=2+x2+2x−3

Let I=x2+2x−32x2+5x−11=x2+2x−32(x2+2x−3)+x−5=2+x2+2x−3x−5

Let I=x2+2x−32x2+5x−11=x2+2x−32(x2+2x−3)+x−5=2+x2+2x−3x−5=2+

Let I=x2+2x−32x2+5x−11=x2+2x−32(x2+2x−3)+x−5=2+x2+2x−3x−5=2+(x+3)(x−1)

Let I=x2+2x−32x2+5x−11=x2+2x−32(x2+2x−3)+x−5=2+x2+2x−3x−5=2+(x+3)(x−1)x−5

Let I=x2+2x−32x2+5x−11=x2+2x−32(x2+2x−3)+x−5=2+x2+2x−3x−5=2+(x+3)(x−1)x−5Now

Let I=x2+2x−32x2+5x−11=x2+2x−32(x2+2x−3)+x−5=2+x2+2x−3x−5=2+(x+3)(x−1)x−5Now(x+3)(x−1)

Let I=x2+2x−32x2+5x−11=x2+2x−32(x2+2x−3)+x−5=2+x2+2x−3x−5=2+(x+3)(x−1)x−5Now(x+3)(x−1)x−5

Let I=x2+2x−32x2+5x−11=x2+2x−32(x2+2x−3)+x−5=2+x2+2x−3x−5=2+(x+3)(x−1)x−5Now(x+3)(x−1)x−5=

Let I=x2+2x−32x2+5x−11=x2+2x−32(x2+2x−3)+x−5=2+x2+2x−3x−5=2+(x+3)(x−1)x−5Now(x+3)(x−1)x−5=(x−1)

Let I=x2+2x−32x2+5x−11=x2+2x−32(x2+2x−3)+x−5=2+x2+2x−3x−5=2+(x+3)(x−1)x−5Now(x+3)(x−1)x−5=(x−1)A

Let I=x2+2x−32x2+5x−11=x2+2x−32(x2+2x−3)+x−5=2+x2+2x−3x−5=2+(x+3)(x−1)x−5Now(x+3)(x−1)x−5=(x−1)A+

Let I=x2+2x−32x2+5x−11=x2+2x−32(x2+2x−3)+x−5=2+x2+2x−3x−5=2+(x+3)(x−1)x−5Now(x+3)(x−1)x−5=(x−1)A+(x+3)

Let I=x2+2x−32x2+5x−11=x2+2x−32(x2+2x−3)+x−5=2+x2+2x−3x−5=2+(x+3)(x−1)x−5Now(x+3)(x−1)x−5=(x−1)A+(x+3)B

Let I=x2+2x−32x2+5x−11=x2+2x−32(x2+2x−3)+x−5=2+x2+2x−3x−5=2+(x+3)(x−1)x−5Now(x+3)(x−1)x−5=(x−1)A+(x+3)B⇒x−5=A(x+3)+B(x−1)

Let I=x2+2x−32x2+5x−11=x2+2x−32(x2+2x−3)+x−5=2+x2+2x−3x−5=2+(x+3)(x−1)x−5Now(x+3)(x−1)x−5=(x−1)A+(x+3)B⇒x−5=A(x+3)+B(x−1)On comparing coefficients we get

Let I=x2+2x−32x2+5x−11=x2+2x−32(x2+2x−3)+x−5=2+x2+2x−3x−5=2+(x+3)(x−1)x−5Now(x+3)(x−1)x−5=(x−1)A+(x+3)B⇒x−5=A(x+3)+B(x−1)On comparing coefficients we get1=A+B,−5=3A−B

Let I=x2+2x−32x2+5x−11=x2+2x−32(x2+2x−3)+x−5=2+x2+2x−3x−5=2+(x+3)(x−1)x−5Now(x+3)(x−1)x−5=(x−1)A+(x+3)B⇒x−5=A(x+3)+B(x−1)On comparing coefficients we get1=A+B,−5=3A−B⇒A=−1,B=2

Let I=x2+2x−32x2+5x−11=x2+2x−32(x2+2x−3)+x−5=2+x2+2x−3x−5=2+(x+3)(x−1)x−5Now(x+3)(x−1)x−5=(x−1)A+(x+3)B⇒x−5=A(x+3)+B(x−1)On comparing coefficients we get1=A+B,−5=3A−B⇒A=−1,B=2Therefore

Let I=x2+2x−32x2+5x−11=x2+2x−32(x2+2x−3)+x−5=2+x2+2x−3x−5=2+(x+3)(x−1)x−5Now(x+3)(x−1)x−5=(x−1)A+(x+3)B⇒x−5=A(x+3)+B(x−1)On comparing coefficients we get1=A+B,−5=3A−B⇒A=−1,B=2ThereforeI=2−

Let I=x2+2x−32x2+5x−11=x2+2x−32(x2+2x−3)+x−5=2+x2+2x−3x−5=2+(x+3)(x−1)x−5Now(x+3)(x−1)x−5=(x−1)A+(x+3)B⇒x−5=A(x+3)+B(x−1)On comparing coefficients we get1=A+B,−5=3A−B⇒A=−1,B=2ThereforeI=2−x−1

Let I=x2+2x−32x2+5x−11=x2+2x−32(x2+2x−3)+x−5=2+x2+2x−3x−5=2+(x+3)(x−1)x−5Now(x+3)(x−1)x−5=(x−1)A+(x+3)B⇒x−5=A(x+3)+B(x−1)On comparing coefficients we get1=A+B,−5=3A−B⇒A=−1,B=2ThereforeI=2−x−11

Let I=x2+2x−32x2+5x−11=x2+2x−32(x2+2x−3)+x−5=2+x2+2x−3x−5=2+(x+3)(x−1)x−5Now(x+3)(x−1)x−5=(x−1)A+(x+3)B⇒x−5=A(x+3)+B(x−1)On comparing coefficients we get1=A+B,−5=3A−B⇒A=−1,B=2ThereforeI=2−x−11+

Let I=x2+2x−32x2+5x−11=x2+2x−32(x2+2x−3)+x−5=2+x2+2x−3x−5=2+(x+3)(x−1)x−5Now(x+3)(x−1)x−5=(x−1)A+(x+3)B⇒x−5=A(x+3)+B(x−1)On comparing coefficients we get1=A+B,−5=3A−B⇒A=−1,B=2ThereforeI=2−x−11+x+3

Let I=x2+2x−32x2+5x−11=x2+2x−32(x2+2x−3)+x−5=2+x2+2x−3x−5=2+(x+3)(x−1)x−5Now(x+3)(x−1)x−5=(x−1)A+(x+3)B⇒x−5=A(x+3)+B(x−1)On comparing coefficients we get1=A+B,−5=3A−B⇒A=−1,B=2ThereforeI=2−x−11+x+32

Step-by-step explanation:

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Answer 2

Answer:

Let I=

x

2

+2x−3

2x

2

+5x−11

=

x

2

+2x−3

2(x

2

+2x−3)+x−5

=2+

x

2

+2x−3

x−5

=2+

(x+3)(x−1)

x−5

Now

(x+3)(x−1)

x−5

=

(x−1)

A

+

(x+3)

B

⇒x−5=A(x+3)+B(x−1)

On comparing coefficients we get

1=A+B,−5=3A−B

⇒A=−1,B=2

Therefore

I=2−

x−1

1

+

x+3

2