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## Main Question or Discussion Point

Hello,

I found something surprising (at least to me) while looking at the following integral:

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

Wolfram Alpha suggests the following substitution as the first step:

[tex]u = \frac{1}{e^x+1}[/tex]

Which leads to the following integral:

[tex]\int \frac{\sqrt{1-2u}}{(u-1)u} du[/tex]

The next substitution is:

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

From there, the solution is just a partial fraction decomposition away. Thinking that one should be able to do both substitutions in one step, I plugged [itex]u[/itex] into [itex]s[/itex] and got:

[tex]s = \sqrt{1-\frac{2}{e^x+1}} = \sqrt{\frac{e^x-1}{e^x+1}}[/tex]

So, this is the

Thank you!

Best regards,

David

I found something surprising (at least to me) while looking at the following integral:

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

Wolfram Alpha suggests the following substitution as the first step:

[tex]u = \frac{1}{e^x+1}[/tex]

Which leads to the following integral:

[tex]\int \frac{\sqrt{1-2u}}{(u-1)u} du[/tex]

The next substitution is:

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

From there, the solution is just a partial fraction decomposition away. Thinking that one should be able to do both substitutions in one step, I plugged [itex]u[/itex] into [itex]s[/itex] and got:

[tex]s = \sqrt{1-\frac{2}{e^x+1}} = \sqrt{\frac{e^x-1}{e^x+1}}[/tex]

So, this is the

**original integrand**that was substituted here! I don't remember having learned this technique. Does this approach of substituting the whole integrand have a name?Thank you!

Best regards,

David