An interesting integral $\int_{0}^{\infty}\frac{dx}{(x+1)(x+2)^a}$ for $a>0$ I want to know how to solve this integral. For $a>0$,
$$I(a)=\int_{0}^{\infty}\frac{dx}{(x+1)(x+2)^a}$$
I tried the substitution $1-u=x+2$, in the hope that I could put the integral in terms of the incomplete Beta function:
$$I(a)=-\int_{-\infty}^{-1}\frac{du}{u(1-u)^a}$$
Which didn't really work out, because the bounds are all wrong. 
I do not know how to proceed. Please help
 A: 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 $2^{-a}\,\Phi\left(\frac{1}{2},1,a\right)$.
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
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\begin{align}
\left.\vphantom{\large A}\mrm{I}\pars{a}\right\vert_{\ \Re\pars{a}\ >\ 0} & \equiv \int_{0}^{\infty}{\dd x \over \pars{x + 1}\pars{x + 2}^{a}}
\,\,\,\stackrel{x + 1\ \mapsto\ x}{=}\,\,\,
\int_{1}^{\infty}x^{-1}\pars{1 + x}^{-a}\,\dd x
\\[5mm] & \stackrel{x\ \mapsto\ 1/x}{=}\,\,\,
\int_{0}^{1}x^{a - 1}\,\pars{1 - x}^{0}\,\pars{1 + x}^{-a}\,\dd x
\\[5mm] & =
\bbx{\,\mrm{B}\pars{a,4}
{}_{2}\mrm{F}_{1}\pars{a,a;a + 1;-1}}
\\[2mm] & \mbox{where}\quad
\pars{\begin{array}{l}
\mrm{B}:\ Beta\ Function
\\
{}_{2}\mrm{F}_{1}:\ Hypergeometric\ Function
\end{array}}
\end{align}
See Euler Type Hypergeometric Function.
