Integrate :$$ \ \int \frac{dv}{4v+\ln v}$$
I have tried in the following way- $ \ \\ Let \ \ \ln v=z , \\ or, dv=vdz , \\ or, dv=e^{z} dz $ $$ $$ Hence, $$ \ \int \frac{dv}{4v+\ln v} = \int \frac{e^{z}}{4e^{z}+z} dz = \frac{1}{4} \int \frac{4e^{z}}{4e^{z}+z}dz = \frac{1}{4} \int \frac{4e^{z}+1-1}{4e^{z}+z}dz$$
$$=\frac{1}{4}\int \frac{4e^{z}+1}{4e^{z}+z}-\frac{1}{4} \int \frac{dz}{4e^{z}+z} \\ =\frac{1}{4} \ln (4e^{z}+z)+c-\frac{1}{4}\int \frac{dz}{4e^{z}+z}.$$
But how can I evaluate the second integral?
Any help is appreciated.
By setting $v=e^t$ the given integral turns into $$ \int \frac{e^t}{4e^t+t}\,dt =\frac{1}{4}\int\frac{dt}{1+\frac{1}{4}t e^{-t}}\tag{1}$$ If the integration range is a subset of $\mathbb{R}^+$ we are allowed to perform a series expansion of the integrand function and apply termwise integration ($\frac{1}{4}te^{-t}$ is positive and bounded by $\frac{1}{4e}$ on $\mathbb{R}^+$), getting
$$ \sum_{n\geq 0}\frac{(-1)^n}{4^{n+1}}\int t^n e^{-nt}\,dt \tag{2}$$ If the integration range equals $\mathbb{R}^+$, $(2)$ turns into $$ \sum_{n\geq 0}\frac{(-1)^n n!}{(4n)^{n+1}}\tag{3} $$ that is a strange-looking but fast-convergent series. The numerical evaluation is simple, but I am pretty sure a simple closed form for $(1)$ does not exist.