Integration by partial fractions?

James2
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Whoa, this here is kicking me hard! Okay, so I've got everything pretty well down until... stuff like... \int \frac{3x + 32}{x^{2}-16x + 64}dx

So, I get how to factor the denominator, but then what? The above won't factor... Also, I read that if the degree of the numerator is higher than the denominator I got to do polynomial long division... I need a review of polynomial long division; Lol.
 
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James2 said:
Whoa, this here is kicking me hard! Okay, so I've got everything pretty well down until... stuff like... \int \frac{3x + 32}{x^{2}-16x + 64}dx

So, I get how to factor the denominator, but then what? The above won't factor... Also, I read that if the degree of the numerator is higher than the denominator I got to do polynomial long division... I need a review of polynomial long division; Lol.
But the numerator is not higher order than the denominator and:$$\frac{3x + 32}{x^{2}-16x + 64}=\frac{3x+32}{(x-8)^2}$$See also:
http://www.math.ucdavis.edu/~kouba/CalcTwoDIRECTORY/partialfracdirectory/PartialFrac.html
 
x^2-16x+64=(x-8)^2
3x+32=3(x-8)+56
 
... oh yes, and complete the square in the numerator.
Thanks lurflurf. The example does not seem to illustrate the following comments does it?
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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