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

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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.
 

Thread 'Prove that the integral is equal to ##\pi^2/8##'

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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