Integration using an Abel transform

[|mathematix|]

Homework Statement

Find the following integral:

Homework Equations

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

The Attempt at a Solution

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

[tex]\sqrt{1-u^{2}}'=-\frac{u}{\sqrt{1-u^2}} \therefore v^{2}=\frac{u^{2}}{1-u^{2}}[/tex]

[tex]du=\frac{dv}{\sqrt{1-u^{2}}}[/tex]

The integral becomes: [tex]\int \frac{dv}{1-u^{4}}[/tex]

I need to somehow get rid off the u and get the integral in terms of v so how can I do that?

 

[haruspex]

The integral becomes: [tex]\int \frac{dv}{1-u^{4}}[/tex]

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?

 

[|mathematix|]

u² = 1 - v², no?

How do you get that?

 

[haruspex]

Maybe I misunderstood your substitutions. This doesn't seem to be consistent:

[tex]u=\sqrt{1-u^{2}}'[/tex]

[tex]\sqrt{1-u^{2}}'=-\frac{u}{\sqrt{1-u^2}}[/tex]

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

 

[|mathematix|]

Maybe I misunderstood your substitutions. This doesn't seem to be consistent:

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

Thanks :)