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Step-by-Step Solution
Step 1: Rewrite the integral in simpler exponential form
The given integral is
$$
\int \frac{e^{3 \log_e (2x)} + 5\, e^{2 \log_e (2x)}}{e^{4 \log_e (x)} + 5\, e^{3 \log_e (x)} - 7\, e^{2 \log_e (x)}} \, dx, \quad x > 0.
$$
Recall that $e^{\log_e (a)} = a$. Hence:
$e^{3 \log_e (2x)} = (2x)^3 = 8 x^3,$
$e^{2 \log_e (2x)} = (2x)^2 = 4 x^2,$
$e^{4 \log_e (x)} = x^4,$
$e^{3 \log_e (x)} = x^3,$
$e^{2 \log_e (x)} = x^2.$
Step 2: Simplify the integrand
Substituting these into the integral gives:
$$
\int \frac{8 x^3 + 5 \cdot (4 x^2)}{x^4 + 5 x^3 - 7 x^2} \, dx
= \int \frac{8 x^3 + 20 x^2}{x^4 + 5 x^3 - 7 x^2} \, dx.
$$
Factor out $x^2$ from numerator and denominator (since $x > 0$, $x \neq 0$):
$$
= \int \frac{x^2(8x + 20)}{x^2(x^2 + 5x - 7)} \, dx
= \int \frac{8x + 20}{x^2 + 5x - 7} \, dx.
$$
Notice that $8x + 20$ can be factored as $4(2x + 5)$, so
$$
\int \frac{8x + 20}{x^2 + 5x - 7} \, dx
= \int \frac{4\,(2x + 5)}{x^2 + 5x - 7} \, dx.
$$
Step 3: Use the substitution
Let
$$
t = x^2 + 5x - 7.
$$
Then
$$
\frac{dt}{dx} = 2x + 5
\quad\Longrightarrow\quad
(2x + 5)\,dx = dt.
$$
Thus the integral becomes:
$$
\int \frac{4 \, (2x + 5)}{x^2 + 5x - 7} \, dx
= \int \frac{4 \, dt}{t}
= 4 \int \frac{dt}{t}.
$$
Step 4: Integrate and substitute back
The integral of $1/t$ with respect to $t$ is $\ln|t|$. Hence,
$$
4 \int \frac{dt}{t} = 4 \ln|t| + C
= 4 \ln|x^2 + 5x - 7| + C.
$$
Therefore, the value of the integral is
$$
\boxed{4\,\ln\bigl|x^2 + 5x - 7\bigr| + C.}
$$
Step 5: Identify the correct option
Comparing with the given options, we see that
$4\;\log_e|x^2 + 5x - 7| + c
\quad \text{matches the correct choice.}
$
