Integration using an Abel transform

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The integral under discussion is transformed using the substitution u = e^x, leading to a new form that simplifies the expression. The attempt involves applying the Abel transform, but confusion arises regarding the substitutions made for u and v. Participants seek clarification on the correct relationships between u and v, particularly whether u^2 = 1 - v^2 is accurate. The conversation highlights the need for consistency in variable substitutions to properly evaluate the integral. Overall, the focus remains on correctly applying the Abel transform to solve the integral.
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Homework Statement



Find the following integral:

Homework Equations



\int \frac{e^{x}}{\sqrt{(1+e^{2x})(1-e^{4x})}}dx

The Attempt at a Solution



I changed the integral to: \int \frac{e^{x}}{(1+e^{2x})\sqrt{(1-e^{2x})}}dx
The let u=e^x
The integral becomes: \int \frac{du}{(1+u^{2})\sqrt{(1-u^{2})}}
I can do this the long way, such as on wolfram alpha but I want to use an Abel transform so let u=\sqrt{1-u^{2}}'

\sqrt{1-u^{2}}'=-\frac{u}{\sqrt{1-u^2}} \therefore v^{2}=\frac{u^{2}}{1-u^{2}}

du=\frac{dv}{\sqrt{1-u^{2}}}

The integral becomes: \int \frac{dv}{1-u^{4}}

I need to somehow get rid off the u and get the integral in terms of v so how can I do that?
 
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|mathematix| said:
The integral becomes: \int \frac{dv}{1-u^{4}}

I need to somehow get rid off the u and get the integral in terms of v so how can I do that?
u2 = 1 - v2, no?
 
haruspex said:
u2 = 1 - v2, no?

How do you get that?
 
Maybe I misunderstood your substitutions. This doesn't seem to be consistent:
|mathematix| said:
u=\sqrt{1-u^{2}}'

\sqrt{1-u^{2}}'=-\frac{u}{\sqrt{1-u^2}}
Did you mean v=\sqrt{1-u^{2}}'? If so, u2 = v2/(1+v2)
 
haruspex said:
Maybe I misunderstood your substitutions. This doesn't seem to be consistent:

Did you mean v=\sqrt{1-u^{2}}'? If so, u2 = v2/(1+v2)

Thanks :)
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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