# partial fractions integral

• Jun 11th 2008, 08:45 PM
dataspot
partial fractions integral
http://img168.imageshack.us/img168/6682/20703745in5.jpg

I'm getting a slightly different answer: I don't come up with the $\displaystyle x^2$, but I've got everything else. Any ideas what I could have missed?

I went the $\displaystyle 2x^3-4x^2-15x+5=A(x-4)+B(x+2)$ route and got $\displaystyle A=\frac{-1}{2}$ and $\displaystyle B=\frac{3}{2}$.
• Jun 11th 2008, 08:50 PM
dataspot
The steps the book follows confuse me, too. You can see them at calcchat.com|Textbook: Calculus, 8th Edition|Chapter 8|Section 5|Exercise 15.

Such as where that $\displaystyle 2x$ and $\displaystyle 5+x$ came from in step 1.
• Jun 11th 2008, 10:03 PM
mr fantastic
Quote:

Originally Posted by dataspot
http://img168.imageshack.us/img168/6682/20703745in5.jpg

I'm getting a slightly different answer: I don't come up with the $\displaystyle x^2$, but I've got everything else. Any ideas what I could have missed?

I went the $\displaystyle 2x^3-4x^2-15x+5=A(x-4)+B(x+2)$ route and got $\displaystyle A=\frac{-1}{2}$ and $\displaystyle B=\frac{3}{2}$.

Since the degree of the numerator is larger than the degree of the denominator you have to do a long division and divide the quadratic into the cubic before attempting to get a partial fraction decomposition. The result of the long division is

$\displaystyle 2x + \frac{x+5}{x^2 - 2x - 8}$.

Now you can attempt to get a partial fraction decomposition ........ of $\displaystyle \frac{x+5}{x^2 - 2x - 8} = \frac{x + 5}{(x - 4)(x + 2)}$.