Where is my mistake in the evaluation of $\int\;\frac {1}{\sqrt {1-e^{2x}}}dx$?

Question

How to evaluate Integral : $$\int\;\frac {1}{\sqrt {1-e^{2x}}}dx $$

My attempt:

I know that my answer is wrong, but I don't know exactly where the error is. I need help finding out?

$$I = \int\;\frac {1}{\sqrt {1-e^{2x}}}dx$$ let : $$t＝e^{x}\;\implies\;dt＝e^{x}dx\;\implies\;dt＝tdx\implies\;dx＝\textstyle\frac {dt}{t}$$ And therefore : $$\implies\;I＝\int\;\frac {1}{\sqrt {1－t^2}}×\frac {1}{t}dt＝\int\frac {1}{t\sqrt {1-t^2}}dt$$ let : $$t＝\cos\theta\;\implies\;dt＝\sin\theta\;d\theta$$ And therefore : $$\implies\;I＝\int\;\frac {\sin\theta d\theta}{\cos\theta\sqrt {1-\cos^2(\theta)}}＝\int\;\frac {－d\theta}{\cos\theta}\\~\\＝-\ln\frac {1+\sin\theta}{\cos\theta}＝\ln\frac {\cos\theta}{1＋\sin\theta}＋c$$

$$using : \sin(\cos^{-1}(t))＝\sqrt {1－t^2}$$

$$\implies\;I＝\ln\frac {\cos(\cos^{-1}(t))}{1＋\sin(\cos^{-1}(t))}＝\ln\frac t{1＋\sqrt {1－t^2}}\\~\\I＝\ln\frac {e^x}{1＋\sqrt {1-e^{2x}}}＋C$$

Answer 1

First note that your integrand is defined only for $x<0.$

Your final result is correct: $$\begin{align}\left(\ln\frac {e^x}{1＋\sqrt {1-e^{2x}}}\right)'&=\frac{1＋\sqrt {1-e^{2x}}}{e^x}\left(\frac {e^x}{1＋\sqrt {1-e^{2x}}}\right)' \\&=\frac{1＋\sqrt {1-e^{2x}}}{e^x}~~~\frac {e^x(1＋\sqrt {1-e^{2x}})-e^x\frac{-2e^{2x}}{2\sqrt {1-e^{2x}}}}{(1＋\sqrt {1-e^{2x}})^2}\\ &=\frac {1＋\sqrt {1-e^{2x}}+\frac{e^{2x}}{\sqrt {1-e^{2x}}}}{1＋\sqrt {1-e^{2x}}} \\ &=\frac {\sqrt {1-e^{2x}}+1-e^{2x}+e^{2x}}{\sqrt {1-e^{2x}}(1＋\sqrt {1-e^{2x}})} \\ &=\frac1{\sqrt {1-e^{2x}}} \end{align}$$ All your calculations (including $\sin\theta=\sqrt{1-\cos^2\theta}$ and your logarithms, which are taken on positive expressions) are also correct, since your two intermediate changes of variable $$t=e^x,\theta=\cos^{-1}(t),$$ force $\theta$ to be in $(0,\pi/2)$ when $x<0.$

Answer 2

Who says that you answer is incorrect. It's correct dude. :)

We don't take the constant term into consideration when evaluating indefinite integrals. So maybe that's why your answer wasn't matching the prescribed one.

If you're ever confused whether your answer is correct or not then go to this calculator then enter the integral that you wish to evaluate. When it gets solved, click on the check answer button (it is a green checkmark alongside the answer) and enter your own answer. It will tell you whether your expression and the answer calculated by the calculator are equivalent or not.