How Do You Solve the Integral ∫x/(6x-x^2)^3/2 dx Using Trig Substitution?

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    Calc 2 Integral
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1. I need help solving this integral, I've tried using trig substitution, but I'm getting nowhere with it.


∫x/(6x-x^2)^3/2dx



Like I said before, I tried changing the bottom to a radical and trig substituting but it doesn't seem to get me anywhere. Thanks in advanced for any help.
 
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Welcome to PF;
You are trying to evaluate the indefinite integral:
$$\int \frac{x}{(6x-x^2)^{3/2}}\;dx$$

Note: ##(6x-x^2)^{3/2} = x^{3/2}(6-x)^{3/2}## ... probably won't help a lot: try again but complete the square first.
 
Oalvarez said:
1. I need help solving this integral, I've tried using trig substitution, but I'm getting nowhere with it.


∫x/(6x-x^2)^3/2dx



Like I said before, I tried changing the bottom to a radical and trig substituting but it doesn't seem to get me anywhere. Thanks in advanced for any help.
Please show your work.
 
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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