Anyone to solve $ \int_0^{\infty} \Big[ (k+1) + (n-k-1) e^{\alpha x} \Big] ^{-2} \alpha x e^{\alpha x} dx $ I'm trying to solve the following formula, but my calculus is quite unhandsome.
$$
\int_0^{\infty} \Big[ (k+1) + (n-k-1) e^{\alpha x} \Big] ^{-2} \alpha x e^{\alpha x} dx
$$
I tried:
$$
\int_0^{\infty} \Big[ (k+1) + (n-k-1) e^{\alpha x} \Big] ^{-2} \alpha x e^{\alpha x} dx = \int_0^{\infty} \Big[ (k+1) + (n-k-1) e^{y} \Big] ^{-2} (y/\alpha) e^{y} dy
$$
And somewhat felt that it looks related to logistic function.
Can anybody give some hints?
 A: With observing
$$\dfrac{d}{d\alpha}\left(\dfrac{1}{(k+1)+(n-k-1)e^{\alpha x}}\right)=\dfrac{-(n-k-1)xe^{\alpha x}}{\left((k+1)+(n-k-1)e^{\alpha x}\right)^2}$$
then the desire integral is
\begin{align}
I
&=\dfrac{\alpha}{-(n-k-1)}\int_0^\infty\dfrac{-(n-k-1)xe^{\alpha x}}{\left((k+1)+(n-k-1)e^{\alpha x}\right)^2}dx\\
&=\dfrac{\alpha}{-(n-k-1)}\dfrac{d}{d\alpha}\int_0^\infty\dfrac{1}{(k+1)+(n-k-1)e^{\alpha x}}dx\\
&=\dfrac{\alpha}{-(n-k-1)}\dfrac{d}{d\alpha}\int_0^\infty\dfrac{e^{-\alpha x}}{(k+1)e^{-\alpha x}+(n-k-1)}dx\\
&=\dfrac{1}{-\alpha(k+1)(n-k-1)}\ln\dfrac{n-k-1}{n}\end{align}
A: Well, we have (assuming that the integral exists):
$$\mathcal{I}_{\space\text{n}}\left(\text{k},\alpha\right):=\int_0^\infty\frac{\alpha\cdot x\cdot\exp\left(\alpha\cdot x\right)}{\left(1+\text{k}+\exp\left(\alpha\cdot x\right)\cdot\left(\text{n}-\text{k}-1\right)\right)^2}\space\text{d}x\tag1$$
Now, first of all let:
 1. $$\beta_1:=1+\text{k}\tag2$$
 2. $$\beta_2:=\text{n}-\text{k}-1\tag3$$
So, we can rewrite equation $\left(1\right)$ as follows:
$$\mathcal{I}_{\space\text{n}}\left(\text{k},\alpha\right)=\int_0^\infty\frac{\alpha\cdot x\cdot\exp\left(\alpha\cdot x\right)}{\left(\beta_1+\exp\left(\alpha\cdot x\right)\cdot\beta_2\right)^2}\space\text{d}x=\alpha\cdot\int_0^\infty\frac{x\cdot\exp\left(\alpha\cdot x\right)}{\left(\beta_1+\exp\left(\alpha\cdot x\right)\cdot\beta_2\right)^2}\space\text{d}x\tag4$$
Now, substitute $\text{u}:=\exp\left(\alpha\cdot x\right)$, so we get:
$$\mathcal{I}_{\space\text{n}}\left(\text{k},\alpha\right)=\alpha\cdot\frac{1}{\alpha^2}\cdot\lim_{\text{p}\to\infty}\int_1^{\exp\left(\alpha\cdot\text{p}\right)}\frac{\ln\left(\text{u}\right)}{\left(\beta_1+\text{u}\cdot\beta_2\right)^2}\space\text{d}\text{u}\tag5$$
Using IBP:
$$\mathcal{I}_{\space\text{n}}\left(\text{k},\alpha\right)=\frac{1}{\alpha\cdot\beta_2}\cdot\left\{\text{Z}+\lim_{\text{p}\to\infty}\int_1^{\exp\left(\alpha\cdot\text{p}\right)}\frac{1}{\text{u}\cdot\left(\beta_1+\text{u}\cdot\beta_2\right)}\space\text{d}\text{u}\right\}\tag6$$
Where:
$$\text{Z}:=\lim_{\text{p}\to\infty}\left[-\frac{\ln\left(\text{u}\right)}{\beta_1+\text{u}\cdot\beta_2}\right]_1^{\exp\left(\alpha\cdot\text{p}\right)}\tag7$$
Assuming that $\alpha\in\mathbb{R}$, we can write:
$$\text{Z}=-\lim_{\text{p}\to\infty}\frac{\alpha\cdot\text{p}}{\beta_1+\exp\left(\alpha\cdot\text{p}\right)\cdot\beta_2}=0\tag8$$
So, we can rewrite equation $\left(6\right)$ as follows:
$$\mathcal{I}_{\space\text{n}}\left(\text{k},\alpha\right)=\frac{1}{\alpha\cdot\beta_2}\lim_{\text{p}\to\infty}\int_1^{\exp\left(\alpha\cdot\text{p}\right)}\frac{1}{\text{u}\cdot\left(\beta_1+\text{u}\cdot\beta_2\right)}\space\text{d}\text{u}=$$
$$\frac{1}{\alpha\cdot\beta_1\cdot\beta_2}\cdot\lim_{\text{p}\to\infty}\left[\ln\left|\frac{\text{u}}{\beta_1\cdot\left(\beta_1+\text{u}\cdot\beta_2\right)}\right|\right]_1^{\exp\left(\alpha\cdot\text{p}\right)}=$$
$$\frac{1}{\alpha\cdot\beta_1\cdot\beta_2}\cdot\lim_{\text{p}\to\infty}\left\{\ln\left|\frac{\exp\left(\alpha\cdot\text{p}\right)}{\beta_1\cdot\left(\beta_1+\exp\left(\alpha\cdot\text{p}\right)\cdot\beta_2\right)}\right|-\ln\left|\frac{1}{\beta_1\cdot\left(\beta_1+\beta_2\right)}\right|\right\}=$$
$$\frac{1}{\alpha\cdot\beta_1\cdot\beta_2}\cdot\lim_{\text{p}\to\infty}\ln\left|\frac{\left(\beta_1+\beta_2\right)\cdot\exp\left(\alpha\cdot\text{p}\right)}{\beta_1+\exp\left(\alpha\cdot\text{p}\right)\cdot\beta_2}\right|\tag9$$
