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?
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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