# Thread:

1. ## Another Integration by Partial Fractions

Hello,

$\int\frac{-x^2-x-1}{5x^2+7x-6}dx$ =

$<br /> \int[\frac{A}{5x-3} + \frac{B}{x+2}]dx$

and

$<br /> B(5x-3) + A(x+2) = -x^2-x-1$

I'm having trouble solving $A$ and $B$. Is there a method for determining them? Or should I be approaching the problem differently altogether?

Austin Martin

2. Originally Posted by auslmar
Hello,

$\int\frac{-x^2-x-1}{5x^2+7x-6}dx$ =

$<br /> \int[\frac{A}{5x-3} + \frac{B}{x+2}]dx$

and

$<br /> B(5x-3) + A(x+2) = -x^2-x-1$

I'm having trouble solving $A$ and $B$. Is there a method for determining them? Or should I be approaching the problem differently altogether?

Austin Martin
Since the degree of the numberator and the denominator are the same you need to use long division before partial fractions!

Remember that if the degree of the numerator is greater than or equal to the denominator ALWAYS use long division.

I don't know how to write out long division with LaTex but

$\frac{-x^2-x-1}{5x^2+7x-6}=-\frac{1}{5}+\frac{\frac{2}{5}x-\frac{11}{5}}{5x^2+7x-6}=-\frac{1}{5}+\frac{1}{5}\left( \frac{2x- 11}{5x^2+7x-6} \right)$

Now we can use partial fractions on $\left( \frac{2x- 11}{5x^2+7x-6} \right)$

Good luck

3. Originally Posted by TheEmptySet
Since the degree of the numberator and the denominator are the same you need to use long division before partial fractions!

Remember that if the degree of the numerator is greater than or equal to the denominator ALWAYS use long division.

I don't know how to write out long division with LaTex but

$\frac{-x^2-x-1}{5x^2+7x-6}=-\frac{1}{5}+\frac{\frac{2}{5}x-\frac{11}{5}}{5x^2+7x-6}=-\frac{1}{5}+\frac{1}{5}\left( \frac{2x- 11}{5x^2+7x-6} \right)$

Now we can use partial fractions on $\left( \frac{2x- 11}{5x^2+7x-6} \right)$

Good luck
Ah yes! I completely forgot about the long division bit. Thanks for the help . I'll work it out!
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