Explanation of integral solution

[justkitten]

I am a bit confused on the solution of this integral.
[MATH]\int\frac{e^x}{1-e^{2x}}[/MATH]The solution I got was: [MATH]\frac{1}{2}ln|\frac{1+e^x}{1-e^x}|+C[/MATH] however, apparently the answer can also be [MATH]tanh^-1(e^x)+C[/MATH]I understand that [MATH]\int\frac{1}{1-u^2}[/MATH] = [MATH]tanh^1u + C [/MATH] if [MATH] |u| < 1[/MATH] and [MATH]\int\frac{1}{1-u^2}[/MATH] = [MATH] coth^1u+ C[/MATH] if [MATH] |u| > 1[/MATH]I am mostly confused on how to determine that |u| is < 1 or > 1 if that makes sense. Is the [MATH]tanh^1[/MATH] used because [MATH]\frac{1}{e^x}[/MATH] is < 1?

 

[Dr.Peterson]

I am a bit confused on the solution of this integral.
[MATH]\int\frac{e^x}{1-e^{2x}}[/MATH]The solution I got was: [MATH]\frac{1}{2}ln|\frac{1+e^x}{1-e^x}|+C[/MATH] however, apparently the answer can also be [MATH]tanh^{-1}(e^x)+C[/MATH]I understand that [MATH]\int\frac{1}{1-u^2}[/MATH] = [MATH]tanh^{-1}u + C [/MATH] if [MATH] |u| < 1[/MATH] and [MATH]\int\frac{1}{1-u^2}[/MATH] = [MATH] coth^{-1}u+ C[/MATH] if [MATH] |u| > 1[/MATH]I am mostly confused on how to determine that |u| is < 1 or > 1 if that makes sense. Is the [MATH]tanh^{-1}[/MATH] used because [MATH]\frac{1}{e^x}[/MATH] is < 1?

You say "apparently". I take that to mean you have seen this stated somewhere without qualification. Is it possible that that source is incomplete, and just failed to mention that this applies only for [MATH]|e^x|<1[/MATH], i.e. for [MATH]x<0[/MATH]? You don't have to assume that they gave their answer with full awareness ...

I see that Wolfram Alpha gives that answer, and also gives a logarithmic version without the absolute value. Both of these are half-truths. Don't trust WA fully, if that is your source!

The form you got is in fact equal to both [MATH]\tanh^{-1}(e^x)[/MATH] and [MATH]\coth^{-1}(e^x)[/MATH] on their respective domains. It will be instructive to graph all three and see what they look like.

In a particular application, such as a definite integral, you would know whether x is positive or negative, and therefore which form is appropriate. Or, you could use [MATH]\coth^{-1}(e^{|x|})[/MATH].

 

[Steven G]

Please do not leave out dx and du from your integrals. When you make substitutions it will hurt you.

 

[justkitten]

In a particular application, such as a definite integral, you would know whether x is positive or negative, and therefore which form is appropriate. Or, you could use [MATH]\coth^{-1}(e^{|x|})[/MATH].

Thank you, this pretty much answered my question.

Also, I am using the James Stewart Calculus textbook and the solution is from there, but yes they did not give any further qualification. I have never used Wolfram Alpha as I tend to avoid the online calculators, but thanks for the tip!

 

[justkitten]

Please do not leave out dx and du from your integrals. When you make substitutions it will hurt you.

Understood! Thank you.

 

[Dr.Peterson]

Thank you, this pretty much answered my question.

Also, I am using the James Stewart Calculus textbook and the solution is from there, but yes they did not give any further qualification. I have never used Wolfram Alpha as I tend to avoid the online calculators, but thanks for the tip!

Interesting. Stewart is quite reputable, so I'm surprised that it would give an incomplete answer (or pose the problem without setting it up to raise the relevant issues). If I were helping you face to face, I'd be looking through the pages before the problem to see if there's a warrant in context to ignore such details.