We are asked to evaluate the integral: $\int \frac{e^{2x}}{e^{3x}-8} dx$

1. Problem Description

We are asked to evaluate the integral:

$\int \frac{e^{2x}}{e^{3x}-8} dx$

2. Solution Steps

Let $u = e^x$. Then $u^2 = e^{2x}$ and $u^3 = e^{3x}$. Also, $du = e^x dx$, so $dx = \frac{du}{e^x} = \frac{du}{u}$.

Substituting these into the integral, we have

$\int \frac{e^{2x}}{e^{3x}-8} dx = \int \frac{u^2}{u^3-8} \frac{du}{u} = \int \frac{u}{u^3-8} du$.

Now, let $u^3 - 8 = (u-2)(u^2 + 2u + 4)$.

So, we have to evaluate $\int \frac{u}{(u-2)(u^2 + 2u + 4)} du$.

We use partial fraction decomposition:

$\frac{u}{(u-2)(u^2+2u+4)} = \frac{A}{u-2} + \frac{Bu+C}{u^2+2u+4}$.

Multiplying both sides by $(u-2)(u^2+2u+4)$ yields:

$u = A(u^2+2u+4) + (Bu+C)(u-2)$

$u = Au^2 + 2Au + 4A + Bu^2 - 2Bu + Cu - 2C$

$u = (A+B)u^2 + (2A-2B+C)u + (4A-2C)$.

Equating coefficients, we have the following system of equations:

$A+B = 0$

$2A-2B+C = 1$

$4A-2C = 0$

From the first equation, $B = -A$. From the third equation, $2C = 4A$, so $C = 2A$. Substituting these into the second equation, we get

$2A - 2(-A) + 2A = 1$

$2A + 2A + 2A = 1$

$6A = 1$

$A = \frac{1}{6}$

Then $B = -\frac{1}{6}$ and $C = \frac{2}{6} = \frac{1}{3}$.

Thus, the integral becomes

$\int \frac{\frac{1}{6}}{u-2} + \frac{-\frac{1}{6}u+\frac{1}{3}}{u^2+2u+4} du = \frac{1}{6} \int \frac{1}{u-2} du + \int \frac{-\frac{1}{6}u+\frac{1}{3}}{u^2+2u+4} du$

$= \frac{1}{6} \ln|u-2| + \int \frac{-\frac{1}{6}u+\frac{1}{3}}{u^2+2u+4} du$.

Let $I = \int \frac{-\frac{1}{6}u+\frac{1}{3}}{u^2+2u+4} du = -\frac{1}{6} \int \frac{u-2}{u^2+2u+4} du$.

We want to rewrite the numerator as a multiple of the derivative of the denominator, which is $2u+2$.

$u-2 = k(2u+2) + l$

$u-2 = 2ku + 2k + l$

$2k = 1$, so $k = \frac{1}{2}$.

$2k+l = -2$, so $1+l = -2$, which means $l=-3$.

Therefore, $u-2 = \frac{1}{2}(2u+2) - 3$.

So, $I = -\frac{1}{6} \int \frac{\frac{1}{2}(2u+2)-3}{u^2+2u+4} du = -\frac{1}{12} \int \frac{2u+2}{u^2+2u+4} du + \frac{1}{2} \int \frac{1}{u^2+2u+4} du$

$I = -\frac{1}{12} \ln|u^2+2u+4| + \frac{1}{2} \int \frac{1}{(u+1)^2+3} du$.

For the last integral, we let $v = u+1$, so $dv = du$.

Then $\int \frac{1}{v^2+3} dv = \frac{1}{\sqrt{3}} \arctan\left(\frac{v}{\sqrt{3}}\right) + C = \frac{1}{\sqrt{3}} \arctan\left(\frac{u+1}{\sqrt{3}}\right) + C$.

So $I = -\frac{1}{12} \ln|u^2+2u+4| + \frac{1}{2\sqrt{3}} \arctan\left(\frac{u+1}{\sqrt{3}}\right) + C$.

Thus, the original integral becomes

$\frac{1}{6} \ln|u-2| -\frac{1}{12} \ln|u^2+2u+4| + \frac{1}{2\sqrt{3}} \arctan\left(\frac{u+1}{\sqrt{3}}\right) + C$

$=\frac{1}{6} \ln|e^x-2| -\frac{1}{12} \ln|e^{2x}+2e^x+4| + \frac{1}{2\sqrt{3}} \arctan\left(\frac{e^x+1}{\sqrt{3}}\right) + C$.

3. Final Answer

$\frac{1}{6} \ln|e^x-2| - \frac{1}{12} \ln(e^{2x}+2e^x+4) + \frac{1}{2\sqrt{3}} \arctan\left(\frac{e^x+1}{\sqrt{3}}\right) + C$