# Another Integration by Partial Fractions

• June 23rd 2008, 07:20 AM
auslmar
Another Integration by Partial Fractions
Hello,

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

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

and

$
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
• June 23rd 2008, 07:39 AM
TheEmptySet
Quote:

Originally Posted by auslmar
Hello,

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

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

and

$
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 (Clapping)
• June 23rd 2008, 08:55 AM
auslmar
Quote:

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 (Clapping)

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