$\displaystyle\int_0^\infty\frac{e^{-2x}x}{(1+e^{-x})^4} \mathrm{d}x=\frac16\left(\ln2-\frac14\right)$. But I don't know how my calculation went wrong Let $e^{-x}=:t$,
get $\def\dd{\mathrm{d}}
x=-\ln t,\, \dd x = -\dfrac1t \dd t;\; t\colon1\!\backsim\!0$,
hence
$$
\def\dd{\mathrm{d}}
\begin{align*}
I
&= \int_1^0 \frac{t\ln t}{(1+t)^4}\dd t \\
&= \frac13\int_0^1 t\ln t\,\dd(1+t)^{-3} \\
&= \underbrace{\frac13\left.\frac{t\ln t}{(1+t)^3}\right|_0^1}_{\kern.48em=~0}
  +\frac13\int_1^0 (1+\ln t)(1+t)^{-3}\dd t \\
&= \frac16\int_0^1 (1+\ln t)\,\dd(1+t)^{-2} \\
&= \frac16\,\underbrace{\!\left.\frac{1+\ln t}{(1+t)^2}\right|_0^1}_{=\raise{-0.05em}{\large\frac14}-1{\color{#F03}{-\ln0}}}
  +\frac16\int_1^0 \frac{\dd t}{t(1+t)^2}.
\end{align*}
$$
 (The part marked in $\color{#F03}{\text{red}}$ is exactly the strange and possibly wrong part)
As to $\displaystyle I_1 = \int_1^0 \frac{\mathrm{d}t}{t(1+t)^2}$,
let $t=\dfrac1u$,
get $\def\dd{\mathrm{d}}
u=\dfrac1t,\, \dd t = -u^{-2}\dd u;\; u\colon1\!\backsim\!\infty$,
hence
$$
\def\dd{\mathrm{d}}
\begin{align*}
I_1
&= \int_\infty^1 \frac{u}{(1+u)^2}\dd u \\
&= \int_\infty^1 \frac{(1+u)-1}{(1+u)^2}\dd u \\
&= \int_\infty^1 \frac{\dd(1+u)}{1+u} - \int_\infty^1 \frac{\dd(1+u)}{(1+u)^2} \\
&= \biggl.\ln(1+u)\biggr|_\infty^1 + \left.\frac1{1+u}\right|_\infty^1 \\
&= \ln2 {\color{#F03}{-\ln(1+\infty)}} + \frac12.
\end{align*}
$$
Therefore
$$
\begin{align*}
I
&= \frac16\left(\frac14-1-\ln0 + \ln2-\ln(1+\infty)+\frac12\right) \\
&= \frac16\left(\ln2-\frac14\right) {\color{#F03}{-\frac16\ln(0\cdot\infty)}}.
\end{align*}
$$

Please correct me, thank you!
(And it would be great if there was a faster way to solve it!)
 A: To avoid undefined limits, integrate by parts instead as follows
\begin{align*}
\int_1^0 \frac{t\ln t}{(1+t)^4}\ dt = &\ \frac16 \int_0^1\ln t\,d\left( -\frac{t^2(3+t)}{(1+t)^3}\right) \\=& \ \frac16 \int_0^1 \frac1{1+t} +\frac1{(1+t)^2} -\frac2{(1+t)^3}\ dt\\= &\ \frac16\ln2 -\frac1{24}
\end{align*}
A: As in the comments Don't directly write things like $\ln(0)$ or $\ln(∞)$. They don't make any sense .Instead see that you have $$\frac{1}{6}\lim_{h\to 0^{+}} \bigg(-\frac{\ln(h)}{(1+h)^{2}}-\ln(1+\frac{1}{h})\bigg)$$ and this is just $0$. Note that you get $\lim_{h\to 0^{+}}\ln(1+\frac{1}{h})$ due to your substitution of $\infty$ in $I_{1}$. I have taken the limit to $0$ for easier computation. And your $-\frac{\ln(h)}{(1+h)^2}$ comes from your substitution of $0$ in the first integral. So when the limit is evaluated, you end up with $0$ and you arrive at the desired answer.
Another way to do it is to expand into a series and then use Dominated Convergence Theorem.
$$\frac{-1}{6}\int_{0}^{\infty}\sum_{r=3}^{\infty}(-1)^{r}r(r-1)(r-2)e^{-(r-3)x}e^{-2x}x\,dx $$
$$\frac{-1}{6}\int_{0}^{\infty}\sum_{r=3}^{\infty}(-1)^{r}r(r-1)(r-2)e^{-(r-1)x}x\,dx $$
$$=\frac{-1}{6}\sum_{r=3}^{\infty}(-1)^{r}\frac{r(r-1)(r-2)}{(r-1)^{2}}$$
$$=\frac{-1}{6}\sum_{k=2}^{\infty}(-1)^{k}\frac{(k+1)(k-1)}{k}$$
Now $$\ln(1+x)=\sum_{r=1}^{\infty}\frac{(-1)^{r-1}x^{r}}{r}=x+\sum_{r=2}^{\infty}\frac{(-1)^{r-1}x^{r}}{r}$$
So $$\ln(1+x)-x=\sum_{r=2}^{\infty}\frac{(-1)^{r-1}x^{r}}{r}$$
Now multiply by $x$ on both sides and differentiate to get
$$\frac{x}{1+x}+\ln(1+x)-2x=\sum_{r=2}^{\infty}(-1)^{r}(r+1)\frac{x^{r}}{r}$$
Now divide by $x$ on both sides and differentiate again to get
$$-\frac{1}{(1+x)^{2}}+\frac{1}{x(1+x)}-\frac{-\ln(1+x)}{x^{2}}=\sum_{r=2}^{\infty}(-1)^{r}(r+1)(r-1)\frac{x^{r-1}}{r}$$ .
Now put $x=1$
And adjust accordingly to get to your answer.
PS:- There could be some errors with the sign in some steps , but I have shown you the idea to compute the integral. I hope you'll be able to compute it and finish it on your own now.
A: The problem is the line that says
$$"= \frac16\,\underbrace{\!\left.\frac{1+\ln t}{(1+t)^2}\right|_0^1}_{=\raise{-0.05em}{\large\frac14}-1{\color{#F03}{-\ln0}}}
  +\frac16\int_1^0 \frac{dt}{t(1+t)^2},"$$
because directly plugging in 0 results in an undefined term. In general, the Integration by Parts Theorem says that if $f,g$ are differentiable on $\left[a,b\right]$ with $f',g'$ integrable on $\left[a,b\right]$, then
$$\int_{a}^{b}f'\left(x\right)g\left(x\right)dx\ =\ f\left(b\right)g\left(b\right)-f\left(a\right)g\left(a\right)-\int_{a}^{a}f\left(x\right)g'\left(x\right)dx.$$
Basically, we need each of those three terms to the right of the equal sign to be finite.
But to finish the question, let's take the limit as $\epsilon \to 0$ (when $\epsilon > 0$) of the original integral. More specifically, let's solve
$$\lim_{\epsilon \to 0}\frac{1}{3}\int_{1}^{\epsilon}\left(1+\ln\left(t\right)\right)\left(1+t\right)^{-3}dt.$$
(Note that your work is still correct even before this integral.)
Using your IBP choice, we get the integral to equal
$$-\frac{1}{3}\lim_{\epsilon \to 0}\left(-\frac{1}{2}\frac{1+\ln{(t)}}{(1+t)^2}\Bigg|_\epsilon^1 - \int_{\epsilon}^1-\frac{dt}{2t(t+1)^2}\right).$$
Notice that we aren't distributing the limit symbol to the two terms because we can only do that with finite terms. After all,
$$\lim_{\epsilon \to 0} -\frac{1}{2}\left(\frac{1+\ln\left(\epsilon\right)}{\left(1+\epsilon\right)^{2}}\right) = \infty.$$
So we shall save computing the limit until the end when we finished evaluating and simplifying an antiderivative. We can use partial fractions on the second integral and do some grunt work. Eventually, we'll get the original integral to be
$$\eqalign{
-\lim_{\epsilon \to 0}\left[\frac{x\left(x+2\right)\ln\left(x\right)+x}{6\left(x+1\right)^{2}}-\frac{\ln\left(x+1\right)}{6}\right]_{\epsilon}^{1} &= -\frac{1}{24}+\frac{\ln\left(2\right)}{6} + \lim_{\epsilon \to 0}\frac{\epsilon\left(\epsilon+2\right)\ln\left(\epsilon\right)+\epsilon}{6\left(\epsilon+1\right)^{2}}.
}$$
To show that limit converges to $0$, we can do
$$\eqalign{
\lim_{\epsilon \to 0}\frac{\epsilon\left(\epsilon+2\right)\ln\left(\epsilon\right)+\epsilon}{6\left(\epsilon+1\right)^{2}} &= \frac{\lim_{\epsilon \to 0}\left(\epsilon\left(\epsilon+2\right)\ln\left(\epsilon\right)+\epsilon\right)}{\lim_{\epsilon \to 0}6\left(\epsilon+1\right)^{2}}.
}$$
It's a little more work to use L'Hôpital's Rule on the numerator, but it goes to $0$.
Does that help?
