Calculating the are between 2 functions, by y-axis So, I have these 2 equations:
$$
y_1=e^{2x}\\y_2=-e^{2x}+4
$$
And I need to calculate the area they have limited right of the y-axis
this is how it looks: 
I have converted the formulas to x:
$$
x_1 = \frac{\log(y_1)}{2}\\x_2 = \frac{\log(4-y_2)}{2}\\
$$
Calculated the limits:
$$
\frac{\log(y_1)}{2}= \frac{\log(4-y_2)}{2}\\y = 2 \Rightarrow x = \frac{\log(2)}{2}
$$
And I got to the point of integrating:
$$
\int_0^{\frac{log(2)}{2}}(\frac{\log(4-y)}{2}- \frac{\log(y)}{2})dy = \\=\int_0^{\frac{log(2)}{2}}(\frac{\log(4-y)}{y})dy
$$
And I don't know how to integrate $\log(x)$ can you please help me?

You don't need to inverse the function. Instead do the following:
Find the intersection by solving: $y_1=y_2$
Integrate $y_2-y_1$ from 0 to the intersection. This will give you the area.

Full calculation:
Solving $y_1=y_2$:
$$e^{2x}=-e^{2x}+4$$
$$2e^{2x}=4$$
$$e^{2x}=2$$
$$2x=\ln(2)$$
$$x=\frac12\ln(2)=\ln(\sqrt2)$$
$$y=e^{2x}=e^{2(\frac12\ln(2))}=2$$
Integrating the difference:
$$\int_{0}^{\frac12\ln(2)}(4-e^{2x}-e^{2x})dx=\int_0^{\frac12\ln(2)}4dx-\int_0^{\frac12\ln(2)}2e^{2x}\\
=2\ln(2)-[e^{\ln(2)}-e^{2*0}]=2\ln(2)-2+1=2\ln(2)-1$$
