[SOLVED] Partial fraction in an integral

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Jul 6th 2008, 01:13 PM
redman223

[SOLVED] Partial fraction in an integral

$\int_{1}^{2} \frac {4y^2-7y-12}{y(y+2)(y-3)} ~dx$

So far I have $\frac {A}{y} + \frac {B}{(y+2)} + \frac {C}{(y-3)}$

You have to combine them right and get a common denominator? So I have $4y^2-7y-12 = A(y+2)(y-3) + B(y)(y-3) + C(y)(y+2)$

But what do I do from here? So far, we have only done examples in class with 2 variables, A and B, but not with C. So its throwing me off.
Jul 6th 2008, 01:35 PM
o_O

I'm assuming that the integral is: $\int_{1}^{2} \frac {4y^2-7y-12}{y(y+2)(y-3)} ~d{\color{red}y}$

You've done well. Just multiply the right-hand side a bit more and collect like terms:
$4y^2-7y-12 = A(y+2)(y-3) + B(y)(y-3) + C(y)(y+2)$

$4y^2-7y-12 = Ay^2 - Ay - 6A \: \: + \: \: By^2 - 3By \: \: + \: \: Cy^2+2Cy$

$4y^2-7y-12 = (A + B + C)y^2 + (-A - 3B + 2C)y - 6A$

Equating the coefficients, we see that:
$A + B + C = 4 \qquad -A - 3B + 2C = -7 \qquad -6A = -12$

etc.etc.
Jul 6th 2008, 01:48 PM
Jhevon

Quote:

Originally Posted by redman223

$\int_{1}^{2} \frac {4y^2-7y-12}{y(y+2)(y-3)} ~d{\color{red}y}$

So far I have $\frac {A}{y} + \frac {B}{(y+2)} + \frac {C}{(y-3)}$

You have to combine them right and get a common denominator? So I have $4y^2-7y-12 = A(y+2)(y-3) + B(y)(y-3) + C(y)(y+2)$ .......................................(1)

But what do I do from here? So far, we have only done examples in class with 2 variables, A and B, but not with C. So its throwing me off.

Alternatively.

call your equation (1):

now, plug in y = 0 into (1). this will get rid of B and C and give you an equation to solve for A

then, plug in y = 3 into (1). this will get rid of A and B and give you an equation with C as the only unknown, so you can solve for it

then, plug in y = -2 into (1). this will get rid of A and C and...you know the drill :p