Question

(2x-5) / (x+3) (x+1)² into partial fractions ​

Answer 1

Answer:

ok

Step-by-step explanation:

Answer 2

Step-by-step explanation:

2x-5/(x+3)(x+1)²

2x-5 = A(x+1)² + B(x+1) (x-3)+ C (x+3)

put x=-1

2(-1)-5 = A(-1+1)² + B (-1+1) + C(-1+3)

-2-5 = C (+2)

-7 = 2C

C = -7/2

2(-3)-5 = A(-3+1)²

-6-5 = A(-2)²

-11 = 4A

A = -11/4

comparing constants

-5 = A + 3B + 3C

-5 = -11/4 + 3B+ 3(-7/2)

3B = -5 + 21/2 + 11 /4

3B = -20+42+11/4

3B = 33/4

B = 11/4

Hence ,

2x-5/(x+3)(x+1)² = -11/4(x+3) + 11/4/(x+1) + -7/2/(x+1)²

= -11/4x+3 + 11/4(x+1) - 7/(x+1)²