Integral using residue theorem complex analysis
Herein, we present an approach that exploits the residue theorem. Let $I(a,b,c)$ be the integral given by
$$\bbox[5px,border:2px solid #C0A000]{I(a,b,c)=\int_0^\infty \frac{x^{a-1}}{(x+b)(x+c)}\,dx} \tag 1$$
where $0<a<1$, $b>0$, and $c>0$.
Next, we move to the complex plane and evaluate the closed contour integral $J(a,b,c)$ given by
$$\bbox[5px,border:2px solid #C0A000]{J(a,b,c)=\oint_C \frac{z^{a-1}}{(z+b)(z+c)}\,dz} \tag2$$
where we choose the branch cut that emanates from the branch point at $z=0$ and extends along the positive real axis to $z=\infty$.
In $(2)$, the contour $C$ is the classical "keyhole" contour comprised of $(i)$ the ray from $\epsilon>0$ to $R>\max(b,c)$ on the upper side of the branch cut, $(ii)$ the counter clockwise circular path on which $z=Re^{i\phi}$, from $\phi =0$ to $\phi =2\pi$, $(iii)$ the ray from $R$ to $\epsilon$ on the lower side of the branch cut, and $(iv)$ the clockwise circular path on which $z=\epsilon e^{i\phi}$, from $\phi=2\pi$ to $\phi=0$.
From the residue theorem, we have
$$\begin{align}J(a,b,c)&=2\pi i \text{Res}\left( \frac{z^{a-1}}{b+c},z=-b,-c\right)\\\\ &=2\pi i \left( \frac{(-b)^{a-1}-(-c)^{a-1}}{c-b}\right)\\\\ &=\bbox[5px,border:2px solid #C0A000]{2\pi i e^{i\pi a}\left(\frac{b^{a-1}-c^{a-1}}{b-c}\right)}\tag 3 \end{align}$$
Next, we express $J(a,b,c)$ as the sum of integrals over the four contours that comprise $C$. Proceeding, we write
$$\begin{align} J(a,b,c)&=\int_\epsilon^R \frac{x^{a-1}}{(x+b)(x+c)}\,dx+\int_0^{2\pi}\frac{iR^{a}e^{ia\phi}}{(Re^{i\phi}+b)(Re^{i\phi}+c)}\,d\phi\\\\ &+\int_R^\epsilon \frac{x^{a-1}e^{i2\pi(a-1)}}{(x+b)(x+c)}\,dx+\int_{2\pi}^0\frac{i\epsilon^{a}e^{ia\phi}}{(\epsilon e^{i\phi}+b)(\epsilon e^{i\phi}+c)}\,d\phi \tag 3 \end{align}$$
As $R\to \infty$ and $\epsilon\to0$, the second and fourth integrals on the right-hand side of $(3)$ approach $0$. Therefore, we can write
$$\bbox[5px,border:2px solid #C0A000]{\lim_{\epsilon \to 0,R\to \infty}J(a,b,c)=(1-e^{i2\pi (a-1)})\int_0^\infty \frac{x^{a-1}}{(x+b)(x+c)}\,dx }\tag 4$$
Putting together $(3)$ and $(4)$ yields
$$\bbox[5px,border:2px solid #C0A000]{\int_0^\infty \frac{x^{a-1}}{(x+b)(x+c)}\,dx=-\frac{\pi}{\sin(\pi a)}\,\frac{b^{a-1}-c^{a-1}}{b-c}}$$
The branch point at $x= 0$ tell us to use $x = e^u$ : $$I = \int_0^\infty \frac{x^{a-1}}{(x+b)(x+c)}dx = \int_{-\infty}^\infty \frac{e^{au}}{(e^u +b)(e^u+c)}du$$ Since $e^{2i \pi a}I =\int_{2i\pi-\infty}^{2i\pi+\infty} \frac{e^{au}}{(e^u +b)(e^u+c)}du$ you have with $C_R$ the rectangular contour $-R \to R\to R+2i\pi\to R-2i\pi \to-R$ and assuming $b > 0,c> 0$ : $$I\frac{1-e^{2i \pi a}}{2i\pi} = \lim_{R \to \infty} \frac{1}{2i\pi}\int_{C_R} \frac{e^{au}}{(e^u +b)(e^u+c)}du $$ $$ = Res( \frac{e^{au}}{(e^u +b)(e^u+c)},\log(b)+i\pi)+Res( \frac{e^{au}}{(e^u +b)(e^u+c)},\log(c)+i\pi)$$ $$ = \frac{e^{a(\log(b)+i\pi)}}{e^{\log(b)+i\pi}(e^{\log(b)+i\pi}+c)}+\frac{e^{a(\log(c)+i\pi)}}{e^{\log(c)+i\pi}(e^{\log(c)+i\pi}+b)} = e^{i \pi a}\frac{b^{a-1}-c^{a-1}}{b-c} $$
By analytic continuation it stays true for every $b,c \in \mathbb{C} \setminus (-\infty,0]$
$\newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
$\ds{\int_{0}^{\infty}{x^{a - 1} \over \pars{x + b}\pars{x + c}} \,\dd x:\ ?.\qquad 0 < a <1\,,\quad b >0\,,\quad c>0}$.
$$\bbox[8px,#efe,border:0.1em groove navy]{\mbox{It's interesting to see that the integral can be performed by 'real methods'}}$$
\begin{align} &\color{#f00}{\int_{0}^{\infty}{x^{a - 1} \over \pars{x + b}\pars{x + c}}\,\dd x} = {1 \over c - b}\pars{\int_{0}^{\infty}{x^{a - 1} \over x + b}\,\dd x - \int_{0}^{\infty}{x^{a - 1} \over x + c}\,\dd x}\label{1}\tag{1} \end{align}
- The LHS integral converges whenever $\ds{0 < \Re\pars{a} < \color{#f00}{2}}$ albeit the OP asked for the condition $\ds{0 < a < \color{#f00}{1}}$.
- The RHS integrals converge whenever $\ds{0 < \Re\pars{a} < \color{#f00}{1}}$ which coincides with the OP condition $\ds{0 < a < 1}$ whenever $\ds{a \in {\mathbb R}}$.
- $\bbox[8px,#ffe,border:0.1em groove navy]{\mbox{Then,}\ \eqref{1}\ \mbox{is evaluated with}\ \ds{0 < \Re\pars{a} < \color{#f00}{1}}\ \mbox{which is more general that the above OP condition}}$
In the first integral we make $\ds{x/b \mapsto x}$ while in the second we make $\ds{x/c \mapsto x}$: \begin{align} &\color{#f00}{\int_{0}^{\infty}{x^{a - 1} \over \pars{x + b}\pars{x + c}}\,\dd x} = {b^{a - 1} - c^{a - 1} \over c - b}\int_{0}^{\infty}{x^{a - 1} \over x + 1} \,\dd x\label{2}\tag{2} \end{align}
Now, $\ds{t \equiv {1 \over x + 1}\implies x = {1 \over t} - 1\implies \totald{x}{t} = -\,{1 \over t^{2}}}$: \begin{align} &\color{#f00}{\int_{0}^{\infty}{x^{a - 1} \over \pars{x + b}\pars{x + c}}\,\dd x} = {b^{a - 1} - c^{a - 1} \over c - b}\int_{1}^{0}t\pars{{1 \over t} - 1}^{a - 1} \,\pars{-\,{\dd t \over t^{2}}} \\[5mm] = &\ {b^{a - 1} - c^{a - 1} \over c - b}\int_{0}^{1}t^{-a}\pars{1 - t}^{a - 1}\,\dd t \label{3.a}\tag{3.a} = {b^{a - 1} - c^{a - 1} \over c - b}\, {\Gamma\pars{-a + 1}\Gamma\pars{a} \over \Gamma\pars{1}} \\[5mm] = &\ \color{#f00}{{b^{a - 1} - c^{a - 1} \over c - b}\,{\pi \over \sin\pars{\pi a}}} \label{3.b}\tag{3.b} \end{align}
Note that the integral involved in \eqref{3.a} $\pars{~the\ Beta\ function~}$ converges whenever $$ \Re\pars{-a} > -1\,,\quad\Re\pars{a - 1} > -1\qquad\implies\qquad 0 < \Re\pars{a} < 1 $$ which defines the general condition for the integrals evaluation.