Partial Fractions for Int 2x^2 - 1 / (4x-1)(x^2+1) dx

[quazzimotto]

I'm not sure why I can't solve? Doing good with u-sub and trig but partial fractions just has me by the nads.

original problem: Int 2x^2 - 1 / (4x-1)(x^2+1)

int 2x^2 - 1 / (4x-1)(x^2+1) = A/(4x-1) + Bx+C/(x^2+1)
2x^2 -1 = A(x^2 +1) + (Bx+C)(4x-1)

2x^2 -1 = Ax^2 + A + 4Bx^2 - Bx + 4Cx - C ok so far?

gathering like terms:

x^2: 2 = A + 4B

x^1: 0 = 4C - 2B

X^0: -1 = A - C

DO I have the equations set up correctly or is there a better way to resolve? Can't get final solutios for A, B C ect....

help appreciated.

 

[stapel]

2x^2 - 1 / (4x-1)(x^2+1)

Does the above mean either of the following?

. . . . .2x² - 1 / [(4x - 1)(x² + 1)]

. . . . .(2x² - 1) / [(4x - 1)(x² + 1)]

Or something else? (I think you mean the latter, but I've learned to ask before assuming....)

Note: Assuming the latter, one method for quickly finding one of the values would be substitution (without first multiplying things out):

. . . . .2x² - 1 = A(x² + 1) + B(4x - 1)

. . . . .Let x = 1/4, so 4x - 1 = 4(1/4) - 1 = 1 - 1 = 0. Then:

. . . . .2(1/4)² - 1 = A((1/4)² + 1) + B(0)

. . . . .2(1/16) - 1 = A(1/16 + 1) + 0

. . . . .1/8 - 1 = A(1/16 + 16/16)

. . . . .1/8 - 8/8 = A(17/16)

. . . . .-7/8 = (17/16)A

. . . . .(-7/8)(16/17) = A

. . . . .(-14/17) = A

Nasty.... :shock:

Eliz.

 

[quazzimotto]

yes - you are correct , wow what a cluster you know what - thanks so much!