Denote
$$I(\alpha, \beta) := \int_0^\infty x^\alpha e^{-\beta x} \log(1 + x) \,dx.$$
A (relatively) nice expression for $I(\alpha, \beta)$ is available for the case that $\alpha$ is a nonnegative integer: Integrating by parts and then substituting $u = 1 + x$ gives
$$I(0, \beta) = \int_0^\infty e^{-\beta x} \log(1 + x) \,dx = \frac1\beta \int_0^\infty \frac{e^{-\beta x} \,dx}{1 + x} = \frac{e^\beta}{\beta} \int_1^\infty \frac{e^{-\beta u} \,du}{u} = \frac{e^\beta}{\beta} \operatorname{E}_1(\beta) ,$$ where $\operatorname{E}_1(x) := \int_x^\infty \frac{e^{-t}}{t} \,dt$ is the exponential integral function. Now, differentiating both sides $\alpha$ times with respect to $\beta$ gives
$$
I(\alpha, \beta)
= (-1)^\alpha \frac{d^\alpha}{d\beta^\alpha} I(0, \beta)
= (-1)^\alpha \frac{d^\alpha}{d\beta^\alpha} \left(\frac{e^\beta}{\beta} \operatorname{E}_1(\beta)\right) .
$$
Direct application of the general Leibniz rule then gives the explicit formula
$$I(\alpha, \beta) = \left[\sum_{k = 0}^\alpha \frac{(-1)^{\alpha + k + 1} {}_\alpha P_k}{\beta^{k + 1}} \right] e^\beta \operatorname{E}_1(\beta) + \sum_{j = 1}^\alpha \sum_{k = 0}^{\alpha - j} \sum_{l = 0}^{j - 1} \frac{(-1)^{\alpha + j + k} {}_{\alpha - j} P_k \cdot {}_{j - 1} P_l}{\beta^{j + k + 2}},$$
where as usual ${}_n P_k = \frac{n!}{(n - k)!}$.
For example,
\begin{align*}
I(1, \beta) &= \left(\frac1{\beta^2} - \frac1\beta\right) e^\beta \operatorname{E}_1(\beta) + \frac{1}{\beta^2} \\
I(2, \beta) &= \left(\frac2{\beta^3} - \frac2{\beta^2} + \frac1\beta\right) e^\beta \operatorname{E}_1(\beta) + \left(\frac3{\beta^3} - \frac1{\beta^2}\right) .
\end{align*}
Somewhat interestingly, if we fix $\beta = 1$, so that $$I(\alpha, 1) = p_\alpha e \operatorname{E}_1(1) + q_\alpha,$$ for integers $p_\alpha, q_\alpha$, then
- $(p_\alpha) = (1, 0, 1, 2, 9, 44, \ldots)$ is the sequence of subfactorials (OEIS A000166), i.e., the number of derangements of $\alpha$ objects, and
- $(q_\alpha) = (0, 1, 2, 8, 36, 200, \ldots)$ appears to be (after the initial $0$) OEIS A233744 (Numbers $p = a(n)$ such that $p$ divided by $(n-1)!$ is equal to the average number of elements of partition sets of $n$ elements excluding sets with a singleton.).
These facts suggest there might be a combinatorial argument for evaluating $I(\alpha, 1)$.
I don't know offhand how to produce it, but Maple gives an unwieldy formula for general $\alpha$:
\begin{multline*}I(\alpha, \beta) \\ = \frac1{\beta^{1 + \alpha}} \Bigg(\left[ (-1)^{-1-\alpha} \pi \alpha \Gamma (\alpha, -\beta) + (-1)^{-\alpha} \pi \Gamma (1 + \alpha) - \frac{\pi e^\beta}{\beta}\right] \csc \pi \alpha \\
\qquad \qquad \qquad - \pi \Gamma (1 + \alpha) \cot \pi \alpha
+ \Psi (-\alpha) \Gamma (1 + \alpha)
- \Gamma (1 + \alpha) \log \beta \Bigg) \\
+ \frac1{\beta^\alpha} {}_2 F_2 (1, 1; 2, 1 - \alpha; \beta) \Gamma(\alpha) .\end{multline*}
Here $\Gamma(\,\cdot\,,\,\cdot\,)$ is the incomplete Gamma function, $\Psi$ is the Digamma function, and ${}_2 F_2$ is a (generalized) hypergeometric function.
Best Answer
By writing $\frac{1}{(x+1)}$ as $\sum_{n\geq 1}\frac{1}{(x+2)^n}$ we have that the original integral equals
$$ \sum_{n\geq 1}\int_{0}^{+\infty}\frac{dx}{(x+2)^{n+a}} = \sum_{m\geq 0}\frac{1}{2^{a+m}(a+m)}$$ i.e. a value of the Lerch trascendent, namely $2^{-a}\,\Phi\left(\frac{1}{2},1,a\right)$.