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

We need one substitution not two which is $e^x-1=\frac{2}{\sqrt{3}}\sec t$. Then, with $e^xdx=\frac{2}{\sqrt{3}}\sec t\tan t dt$ and $2e^x-1=\frac{4}{\sqrt{3}}\sec t+1$ we have $$I=\int\frac{e^x(2e^x-1)}{\sqrt{3(e^x-1)^2-4}}dx=\int\frac{\frac{4}{\sqrt{3}}\sec t+1}{2\tan t}\frac{2}{\sqrt{3}}\sec t\tan t dt=\int (\frac{4}{3}\sec^2 t+\frac{1}{\sqrt{3}}\sec t)dt$$ and $$I=\frac{4}{3}\tan t +\frac{1}{\sqrt{3}}\ln|\sec t+\tan t|+c.$$ in terms of $x$, $$I=\frac{2}{3}\sqrt{3e^{2x}-6e^x-1} +\frac{1}{\sqrt{3}}\ln|\sqrt{3}(e^x-1)+\sqrt{3e^{2x}-6e^x-1}|+c.$$

Finally, https://www.wolframalpha.com/input?i=integral+e%5Ex%282e%5Ex-1%29%2Fsqrt%283e%5E%282x%29-6e%5Ex-1%29 I have to check myself.

Answer 2

I wouldn't split the numerator, but rather complete the square under the root and then turn it into a trigonometric integral with $y=\sqrt{3}u-\sqrt{3}$ and $y=2\sec{v}$. Then use trig identities to simplify and evaluate the integral.

Answer 3

Hint… From where you are, for the first integral, substitute $$t=3u^2-6u-1$$ and for the second integral, complete the square inside the square root and substitute $$y=\sqrt{3}(u-1)$$ to obtain an $\operatorname{arcosh}$ type integral

Answer 4

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