# Thread: partial fractions integral

1. ## partial fractions integral

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

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

2. 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 $2x$ and $5+x$ came from in step 1.

3. Originally Posted by dataspot

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

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

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

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