Integrate with which technique $\int^{\ln6}_{\ln4}e^x\ln(e^{2x}-4)\,dx$  Hey, I have tried about everything, I used parts, , letting $dv=e^{2x}-4$ but it made the solution more complicated.  I believe the solution has something to do with a substitution of $e^{2x}-4$, but $u=e^{2x}-4$ doesn't seem to work because the derivative is $du=2e^{2x}$, which does not make the solution easier as well, I am a little stuck, any help would be appreciated

HINT:
Setting $e^x=y$
$$\int_4^6\ln(y^2-4)\ dy$$
Now $\ln(y^2-4)=\ln(y-2)+\ln(y+2)$
Integrating by Parts,
$$\int \ln(y+a)\ dy=\ln(y+a)\int dy-\int\left[\dfrac{d\{\ln(y+a)\}}{dy}\cdot\int dy\right]\,dy$$
Can you take it home from here?
