Integration of an inverse function

iceman_ch
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Homework Statement



\int\frac{4}{x(x+3)}

Homework Equations





The Attempt at a Solution



I can get to s certain point and I know I need to do substitution but, everytime I try a substitution it just creates a more difficult problem.

4\int(x^{-1}(x+3)^{-1})

I've tried substitution x^-1 for U and using (x+3)^-1 for dv but, none of it works. If someone could give me a gentle nudge it would be appreciated.

Thanks
 
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This problem is an ideal candidate for the method of "partial fractions".

Try decomposing \frac{1}{x(x+3)} into the form \frac{A}{x}+\frac{B}{x+3} where A and B are constants you need to determine.
 
OH man so obvious. Your the man thank you so much. I havn't had a math class in over a year and now I'm taking diff eq. Bad idea you should definately keep them all together.
 
And, when talking about functions, be careful to distinguish between "reciprocal" and "inverse" functions!
 
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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