Please correct me if I am wrong. Definite integrals So I solved this $\displaystyle\int_{\ln3}^{0}\frac{1-e^x}{1+e^x}{\mathrm{d} x}$  and got $-\ln12$ as an answer. Am I right? 
My attempt
$$I=\int_{\ln3}^{0}\frac{1-e^x}{1+e^x}{\mathrm{d} x} = \int_{\ln3}^{0}-\frac {1+e^x}{1-e^x}{\mathrm{d} x}$$
$$I= \int_{\ln3}^{0}-\frac{1+e^x-e^x+e^x}{1-e^x}{\mathrm{d} x}$$
$$I=\int_{\ln3}^{0}-\frac{1-e^x}{1-e^x}{\mathrm{d} x} + \int_{\ln3}^{0}\frac{2e^x}{1-e^x}{\mathrm{d} x}$$
$$I=-\int_{\ln3}^{0}1\mathrm{d} x+2\int_{\ln3}^{0}\frac{e^x}{1-e^x}{\mathrm{d} x}$$
 A: Your idea is good but you messed negative signs. You changed some signs in the integrand.This line is not correct:
$$I=\int_{ln3}^{0}\frac{1-e^x}{1+e^x}{\mathrm{d} x} = \int_{ln3}^{0}\color{red}{-}\frac {1\color{red}{+}e^x}{1\color{red}{-}e^x}{\mathrm{d} x}$$
Your attempt should be :
$$I=\int_{\ln3}^{0}\dfrac{1-e^x}{1+e^x}{\mathrm{d} x} $$
$$I= \int_{\ln3}^{0}\dfrac{1+e^x-e^x-e^x}{1+e^x}dx$$
$$I= \int_{\ln3}^{0}dx+\int_{\ln3}^{0}\frac{-2e^x}{1+e^x}dx$$
A: Hint:
$$\int_{\ln 3}^{0}\frac{1-e^x}{1+e^x}\mathrm dx=\int_{\ln 3}^{0}\left(1-\frac{2e^x}{1+e^x}\right)\mathrm dx$$
Can you proceed? I think you might have made a mistake in solving the logarithm you obtained as the answer comes out to be $\ln(4/3)$. 
A: $$\displaystyle J= \int_{\ln3}^{0}\frac{1-e^x}{1+e^x}{\mathrm{d} x} = \int_0^{\ln 3} \frac{\sinh x/2}{\cosh x/2} \, dx = 2 \left. \ln( \cosh x/2)\right|_0^{\ln 3}=2 \ln\left(\cosh\left(\frac{\ln 3}{2}\right)\right),$$
$$\cosh^2\left( \frac{\ln 3}{2} \right)=\left( \frac{\sqrt{3}+1/\sqrt{3}}{2}   \right)^2=\frac{4}{3}.$$
So $$J=\ln \left(\frac{4}{3}\right)$$
A: I just looked at the work you mentioned in comments.
Check your first step, thats incorrect 
$a+b$ = $-(-a-b)$ and Not $-(a-b)$
