I was trying to solve $I = \int_0^\infty \frac{t}{e^t-1}dt$.
My approach
I took the more general form of integral $f(s) = \int_0^{\infty}\frac{e^{-st}}{e^t-1}dt$ the same way as this post on Math Stack Exchange.
So, my answer $$I = -\left[\frac{\partial f(s)}{\partial s}\right]_{s =0}$$ but that simply doesn't work.
However, $\int_0^\infty \frac{t}{e^t-1}dt$ without using series doesn't answer my question as I'm interested in finding why finding derivative at $s = 0$ doesn't work.
Also, I'm interested in finding a number of different ways I can solve this problem.
Well! This integral is really interesting and has a lot to do with Bernoulli's Numbers, Riemann Zeta function, Polygamma function, and, in the last, what you were trying is nothing less than Laplace Transformation.
$$I = \int_0^{\infty}\frac {t}{e^t-1}dt$$First, I'll try to answer why $$I =-\frac{\partial}{\partial s}\int_0^{\infty}\frac {e^{-st}}{e^t-1}dt$$ is failing to conclude the answer, and after that, I'll try to add a few other ways you can conclude the answer.
Why $I =-\frac{\partial}{\partial s}\int_0^{\infty}\frac {e^{-st}}{e^t-1}dt$ fails
Answer On the very first look, you can conclude that the definite integral $\int_0^{\infty}\frac {e^{-st}}{e^t-1}dt$ doesn't exist (Undefined).
Hint The lower limit of the definite integral is $0$, and when $t = 0$, the numerator = $1$, but the denominator = $0$. However, for $\int_0^{\infty}\frac {t}{e^t-1}dt$, it's totally different. Even if you try to find it by taking the Laplace transformation, you'll eventually end up with $\Gamma(0)$.
How to solve using Transformation only?
This may answer your question, but as you asked for a few other ways to solve this integral, I'm adding a few transformation-based answers.
$1]$ Using Laplace Transform Identity
I'm unable to provide a link from where I found this, but if I could remember, you can find it in "Handbook of Mathematics for Engineer's & Scientists" by Andrei & Alexander.
$$\begin{align*} F(s) = \mathcal{L}f(t) &= \int_0^{\infty}e^{-st}f(t)dt\\ & \implies \sum_{s=1}^{\infty}F(s) = \int_0^{\infty} \frac {f(t)}{e^t-1}dt\\ & \text{We can put } f(t) = t \implies \mathcal{L}(t) = \frac 1{s^2}\\ & \text{Thus,}\\ & \color{green}{I = \int_0^{\infty} \frac{t}{e^t-1}dt = \sum_{s=1}^{\infty}\frac {1}{s^2}=\zeta(2) = \frac {\pi^2}{6}}\\ \end{align*}$$$2]$ Using digamma function
$$\begin{align*} I &= \int_0^{\infty}\frac t{e^t-1}dt\\ & \text{We substitute, } e^{-t} = x \implies t = -\ln x \implies dt = -\frac{dx}{x} \\ & \color{blue}{I = -\int_0^1 \frac {\ln x}{1-x}dx}\\ \end{align*}$$Now,
$$\color{red}{\psi_0 (z) = -\gamma + \int_0^{1} \frac {1-x^{z-1}}{1-x}dx}$$We'll differentiate, and that is trigamma function $\psi_1(z)$.
$$\implies \frac {\partial\psi_0}{\partial z}= \psi_1 (z) = -\int_0^1 \frac {x^{z-1}\ln x}{1-x}dx$$ $$\color{green}{I = \left[\psi_1(z)\right]_{z=1} = \frac {\pi^2}{6}}$$$3]$ Using the integral form of even values of zeta function
$$\left|\zeta(2n) = \frac 1{(2n-1)!}\int_0^{\infty}\frac{x^{2n-1}}{e^x-1}dx\right|_{n =1}$$$4]$ More general form of the above integral in terms of Bernoulli's Number
$$\left|\int_0^{\infty} \frac {x^{2n-1}}{e^{px}-1}dx = (-1)^{n-1}\left(\frac {2\pi}{p}\right)^{2n} \frac{B_{2n}}{4n}\right|_{n, p = 1, 1}$$ $B_2 = \frac 1{6}$.My last attempt: The cosine Transformation
$$f_c(u) = \int_0^{\infty}f(x)\cos(ux)dx$$For $$\text{if }f(x) = \frac x{e^{ax}-1} \implies f_c(u) = \frac 1{2u^2} - \frac {\pi^2}{2a^2\sinh^2(\pi a^{-1}u)}$$
I tried to take the limit as $u \to 0$ but didn't work. Maybe because the cosine transformation is defined for $u \in (0, \infty)$, but at $u = 0.0001$, it's $f_c(u = 0.0001) = 1.644934043288231 \approx \pi^2/6$.