Calculate $\int\frac{2e^{2x}-e^x}{\sqrt{3e^{2x}-6e^x-1}}dx$

Question

The question is to evaluate I, which is defined as follows:
$$
I \equiv \int\frac{2e^{2x}-e^x}{\sqrt{3e^{2x}-6e^x-1}}dx
$$
The first thing I did was simplify the numerator of the integrand by factoring out an $e^x$ to get
$$
\int\frac{e^x(2e^x-1)}{\sqrt{3e^{2x}-6e^x-1}}dx
$$
We then let $u = e^x$, then $dx = e^{-x}du$, and this enables us to rewrite $I$ as follows after some simplification
$$
\int\frac{2u – 1}{\sqrt{3(u^2-2u)-1}}du
$$
We can then split the integrand as follows by rewriting the numerator as $2u – 2 + 1$
$$
\int\frac{2u-2}{\sqrt{3(u^2-2u)-1}}+\frac{1}{\sqrt{3(u^2-2u)-1}}du
$$
We can then apply the sum rule to split the integral as follows
$$
\int\frac{2u-2}{\sqrt{3(u^2-2u)-1}}du + \int\frac{1}{\sqrt{3(u^2-2u)-1}}du
$$
Then factoring out a 2 from the first integral gives
$$
2\int\frac{u-1}{\sqrt{3(u^2-2u)-1}}du + \int\frac{1}{\sqrt{3(u^2-2u)-1}}du
$$
I'm not sure how to solve these integrals from here though, so any help would be great.

Answer 1

The first of your two final integrals is $$\int\frac{u-1}{\sqrt{3(u^2-2u)-1}}du=\frac16\int\frac{d(3(u^2-2u)-1)}{\sqrt{3(u^2-2u)-1}}=\left[\frac13\sqrt{3(u^2-2u)-1}\right].$$

For the second one, let $v=\frac{\sqrt3}2(u-1).$ Then, $$\int\frac{du}{\sqrt{3(u^2-2u)-1}}=\frac1{\sqrt3}\int\frac{dv}{\sqrt{v^2-1}}=\left[\frac1{\sqrt3}\operatorname{arcosh}v\right].$$