justkitten
New member
- Joined
- Sep 11, 2020
- Messages
- 3
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?
[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?