Question

how do i use partial fractions for this ??

Answer 1

\[denominator : (x-1)(x+1)(x^2+4) \]

Answer 2

\[\frac{ 1 }{ (x-1)(x+1)(x^2+4) } = \frac{ A }{ x-1 } + \frac{ B }{ x+1 } + \frac{ C }{ x^2+4 } \]

Answer 3

i am getting A = 1/10 , B = - 1/10 but cannot solve for C

Answer 4

when you have terms like ax^2 +b in the denominator,
the numerator must be Cx+D

try this :
\(\Large \frac{ 1 }{ (x-1)(x+1)(x^2+4) } = \frac{ A }{ x-1 } + \frac{ B }{ x+1 } + \frac{ Cx+D }{ x^2+4 }\)

now try to find A,B,C,D
let me know if you get stuck :)

Answer 5

***must be linear like Cx+D

Answer 6

putting x = -1 , i get B = -1/10
putting x = 1 , i get A = 1/10
how do i solve for C and D (Cx+D) (x-1)(x+1) for x = 1 and -1 it becomes zero
@hartnn

Answer 7

find the common denominator and compare the numerators!

Answer 8

now there are no denominators (x-1)(x+1)(x^2+4) have been cancelled from both LHS and RHS

Answer 9

\(\Large 1 = A (x+1)(x^2+4)+B(x-1)(x^2+4)\\ \Large \quad \: \: \:+(Cx+D)(x+1)(x-1)\)

see if you can get that ^^

Answer 10

yes even i got that when i put x=-1 everything cancels out expect 1= B(-1-1) (1+4)
when i put x =1 everything cancels out expect 1= A(1+1) (1+4)
but what to do for c and d .. can't figure it out

Answer 11

Then you distribute the terms out
like
B(x-1)(x^2+4) = B(x^3 +4x-x^2-4) = Bx^3 +4Bx -Bx^2 -4B

got that ^^?

Answer 12

do it for other 2 terms also

then collect all x^3 terms together
all x^2 terms together and so on

Answer 13

and solve the equations ^_^