partial fraction

if 1/x3(x+2) = A/x+B/x2+C/x3+D/(x+2) then value of |1/A 1/B |
|1/D 1/C |

7 Answers

1
Honey Arora ·

ans is -16

62
Lokesh Verma ·

if 1/x3(x+2) = A/x+B/x2+C/x3+D/(x+2) then value of |1/A 1/B |
|1/D 1/C |

multiply both sides by (x+2)

1/x3 = (x+2)(A/x+B/x2+C/x3)+D

substitute x=-1

-1/8=D
also, multiply both sides by x3

1/(x+2) = Ax2+Bx+C+Dx3/(x+2)

substitute x=0

we get 1/2=C

62
Lokesh Verma ·

for the other parts, balance the pwers of x by multiplying by x3.(x+2)

1
Honey Arora ·

A=1/8
B=-1/4
C=1/2
D=-1/8

1
Honey Arora ·

isn't it easy if we write like this 1=A(x3+2x2)+B(x2+2x)+C(x+2)+Dx3
=> A+D=0
2A+B=0
2B+C=0
and 2C=1

solving these u cld get the values

62
Lokesh Verma ·

yes indeed honey it makes it much easier...

I forgot that method.. :)

Good that you recalled this one :)

1
Honey Arora ·

much more thn u.......we need to recall those

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