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?

To bypass your problem (I dunno how to integrate $\log$ too $\ddot \smile$)
$$\begin{align}
\text{At point of intersection,$y_1=y_2$}&\implies x=\frac{\ln 2}{2}\\
\text{Area}&=\int_{0}^{\large\frac{\ln 2}{2}}\left[{(-e^{2x}+4)-e^{2x}}\right]\ \mathrm dx\\
&=\int_{0}^{\large\frac{\ln 2}{2}}\left({4-2e^{2x}}\right)\ \mathrm dx\\
\end{align}$$
