The associated homogeneous equation is $4f''(x) + 4f'(x) + f(x) = 0$. Its characteristic equation $4r^2 + 4r + 1 = 0$ gives $r = -\frac{1}{2}$ as a repeated root. Thus, the general solution to the homogeneous equation is $f_h(x) = e^{-\frac{x}{2}}(C_1 + C_2 x)$. Using the boundary condition $e^{\frac{x}{2}} f(x) \to 0$, we find $C_1 = 0$ and $C_2 = 0$, so $f_h(x) = 0$. (A more elaborated explanation can be found below.)
For the particular solution, the right-hand side $\frac{1}{e^x - 1}$ can be written as $\sum_{n=1}^\infty e^{-n x}$. Assume $f_p(x) = \sum_{n=1}^\infty A_n e^{-n x}$. Substituting into the differential equation gives $(4n^2 - 4n + 1) A_n = 1$, so $A_n = \frac{1}{(2n-1)^2}$. Thus, the solution is $f(x) = \sum_{n=1}^\infty \frac{e^{-n x}}{(2n-1)^2}$.
To evaluate $\int_0^\infty f(x) \, dx$, we compute:
$$
\int_0^\infty f(x) \, dx = \int_0^\infty \sum_{n=1}^\infty \frac{e^{-n x}}{(2n-1)^2} \, dx = \sum_{n=1}^\infty \frac{1}{(2n-1)^2} \int_0^\infty e^{-n x} \, dx.
$$
The integral $\int_0^\infty e^{-n x} \, dx = \frac{1}{n}$, so:
$$
\int_0^\infty f(x) \, dx = \sum_{n=1}^\infty \frac{1}{n (2n-1)^2}.
$$
Using partial fraction decomposition, $\frac{1}{n(2n-1)^2} = \frac{1}{n} - \frac{2}{2n-1} + \frac{2}{(2n-1)^2}$. Substituting this:
$$
\sum_{n=1}^\infty \frac{1}{n(2n-1)^2} = \sum_{n=1}^\infty \frac{1}{n} - \sum_{n=1}^\infty \frac{2}{2n-1} + \sum_{n=1}^\infty \frac{2}{(2n-1)^2}.
$$
The series $\sum_{n=1}^\infty \frac{1}{n}$ diverges, but in combination with $-\sum_{n=1}^\infty \frac{2}{2n-1}$, the divergent parts cancel, leaving:
$$
\sum_{n=1}^\infty \left( \frac{1}{n} - \frac{2}{2n-1} \right) = -2 \ln 2.
$$
The series $\sum_{n=1}^\infty \frac{2}{(2n-1)^2}$ converges to $\frac{\pi^2}{4}$. Combining these results:
$$
\int_0^\infty f(x) \, dx = -2 \ln 2 + \frac{\pi^2}{4}.
$$
Thus, the value of the integral is:
$$
\boxed{\frac{\pi^2}{4} - 2 \ln 2}
$$
$C_1$ and $C_2$ ARE BOTH ZERO
The homogeneous solution is $f_h(x) = e^{-\frac{x}{2}} (C_1 + C_2 x)$. Multiplying by $e^{\frac{x}{2}}$ gives $e^{\frac{x}{2}} f_h(x) = C_1 + C_2 x$. As $x \to \infty$: if $C_2 \neq 0$, the term $C_2 x$ grows without bound; if $C_2 = 0$ but $C_1 \neq 0$, the term $C_1$ remains constant. In both cases, $e^{\frac{x}{2}} f_h(x)$ does not approach zero unless both $C_1 = 0$ and $C_2 = 0$.
The particular solution is $$f_p(x) = \sum_{n=1}^\infty \frac{e^{-n x}}{(2n - 1)^2}.$$ Multiplying by $e^{\frac{x}{2}}$ yields $$e^{\frac{x}{2}} f_p(x) = \sum_{n=1}^\infty \frac{e^{-(n - \frac{1}{2})x}}{(2n - 1)^2}.$$ For each term in the series, $n - \frac{1}{2} \geq \frac{1}{2}$ since $n \geq 1$, so $e^{-(n - \frac{1}{2})x} \leq e^{-\frac{x}{2}}$. As $x \to \infty$, each term $e^{-(n - \frac{1}{2})x}$ decays exponentially to zero, implying $e^{\frac{x}{2}} f_p(x) \to 0$ as $x \to \infty$.
The general solution is $f(x) = f_h(x) + f_p(x)$. Multiplying by $e^{\frac{x}{2}}$ gives $$e^{\frac{x}{2}} f(x) = e^{\frac{x}{2}} f_h(x) + e^{\frac{x}{2}} f_p(x) = C_1 + C_2 x + e^{\frac{x}{2}} f_p(x).$$ Taking the limit as $x \to \infty$, we get $$\lim_{x \to \infty} e^{\frac{x}{2}} f(x) = \lim_{x \to \infty} (C_1 + C_2 x) + \lim_{x \to \infty} e^{\frac{x}{2}} f_p(x).$$ Since $e^{\frac{x}{2}} f_p(x) \to 0$, we find $$\lim_{x \to \infty} (C_1 + C_2 x) = 0.$$ For this to hold, $C_2$ must be zero; otherwise, $C_2 x$ grows without bound. With $C_2 = 0$, $C_1$ must also be zero to satisfy the condition.
Since neither $C_1$ nor $C_2$ can take non-zero values without violating the boundary condition, we must have $C_1 = 0$ and $C_2 = 0$. This ensures that the homogeneous solution does not contribute to the asymptotic behavior of $f(x)$, and the boundary condition is satisfied solely by the particular solution.
