We need to evaluate the integral $\int \frac{e^x}{e^{3x}-8} dx$.

1. Problem Description

We need to evaluate the integral $\int \frac{e^x}{e^{3x}-8} dx$.

2. Solution Steps

Let $u = e^x$. Then $du = e^x dx$.

The integral becomes

$\int \frac{1}{u^3 - 8} du$.

We can factor the denominator as a difference of cubes:

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

We perform a partial fraction decomposition:

$\frac{1}{u^3 - 8} = \frac{A}{u-2} + \frac{Bu+C}{u^2+2u+4}$.

Multiplying both sides by $u^3 - 8$ gives

$1 = A(u^2+2u+4) + (Bu+C)(u-2)$

$1 = Au^2 + 2Au + 4A + Bu^2 - 2Bu + Cu - 2C$

$1 = (A+B)u^2 + (2A - 2B + C)u + (4A - 2C)$

Comparing coefficients:

$A+B = 0$

$2A - 2B + C = 0$

$4A - 2C = 1$

From the first equation, $B = -A$. Substituting into the second equation gives:

$2A - 2(-A) + C = 0$

$4A + C = 0$

$C = -4A$

Substituting into the third equation:

$4A - 2(-4A) = 1$

$4A + 8A = 1$

$12A = 1$

$A = \frac{1}{12}$

Then $B = -\frac{1}{12}$ and $C = -4(\frac{1}{12}) = -\frac{1}{3}$.

So the integral becomes:

$\int \left( \frac{1/12}{u-2} + \frac{(-1/12)u - 1/3}{u^2+2u+4} \right) du$

$= \frac{1}{12} \int \frac{1}{u-2} du - \frac{1}{12} \int \frac{u+4}{u^2+2u+4} du$

For the second integral, we complete the square in the denominator:

$u^2+2u+4 = (u+1)^2 + 3$.

We want to rewrite the numerator in terms of the derivative of the inside:

$u+4 = \frac{1}{2}(2u+2) + 3$

So $\int \frac{u+4}{u^2+2u+4} du = \int \frac{(1/2)(2u+2) + 3}{u^2+2u+4} du = \frac{1}{2} \int \frac{2u+2}{u^2+2u+4} du + 3\int \frac{1}{(u+1)^2+3} du$

$= \frac{1}{2} \ln(u^2+2u+4) + 3 \cdot \frac{1}{\sqrt{3}} \arctan \left( \frac{u+1}{\sqrt{3}} \right)$

$= \frac{1}{2} \ln(u^2+2u+4) + \sqrt{3} \arctan \left( \frac{u+1}{\sqrt{3}} \right)$

So the original integral becomes:

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

$= \frac{1}{12} \ln|u-2| - \frac{1}{24} \ln(u^2+2u+4) - \frac{\sqrt{3}}{12} \arctan \left( \frac{u+1}{\sqrt{3}} \right) + C$

Substituting back $u = e^x$:

$\frac{1}{12} \ln|e^x-2| - \frac{1}{24} \ln(e^{2x}+2e^x+4) - \frac{\sqrt{3}}{12} \arctan \left( \frac{e^x+1}{\sqrt{3}} \right) + C$

3. Final Answer

$\frac{1}{12} \ln|e^x-2| - \frac{1}{24} \ln(e^{2x}+2e^x+4) - \frac{\sqrt{3}}{12} \arctan \left( \frac{e^x+1}{\sqrt{3}} \right) + C$