partial fractions integral

**dataspot** · June 11th 2008, 08:45 PM

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 $B=\frac{3}{2}$ .

**dataspot** · June 11th 2008, 08:50 PM

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.

**mr fantastic** · June 11th 2008, 10:03 PM

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 $\frac{x+5}{x^2 - 2x - 8} = \frac{x + 5}{(x - 4)(x + 2)}$ .

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