Need help for integration I need help with finding the following integral 
$$\int_0^{+\infty} \frac{x^a\,dx}{(1+x)^2(A+x^a)}, $$ where $A>0$ and $a \in \mathbb{R}\,.$
This is my first trial.
\begin{align} 
\int_0^\infty \dfrac{x^a\,{\rm d}x}{(1+x)^2\,(A+ x^a)} \\
          &= \int_0^1 \dfrac{x^a\,{\rm d}x}{(1+x)^2\,(A+ x^a)} 
        + \int_1^\infty \dfrac{x^a\,{\rm d}x}{(1+x)^2\,(A+ x^a)} =: I_1 + I_2\,. \notag
  \end{align}
After applying binomial expansion and interchanging the infinite summation and integral I got the (possible) solution for $I_1$ as
\begin{align} \label{2.1}
     I_1 &= \int_0^1 \sum_{k \geq 0} \binom{-2}k \dfrac{x^{a+k}\, {\rm d}x}{A+ x^a} 
        = \dfrac1a \sum_{k \geq 0} (-1)^k \dfrac{(2)_k}{k!} \int_0^1 \dfrac{t^{\frac{k+1}a}\, 
        {\rm d}t}{A+t} \notag \\
     &= \dfrac1A \sum_{k \geq 0} (-1)^k \dfrac{(2)_k}{k!\, (a+k+1)}\,
        {}_2F_1\left(1, \frac{k+1}a+1; \frac{k+1}a+2; -\dfrac1A\right)\notag \\
     &= \dfrac1A \sum_{k \geq 0} (-1)^k \dfrac{k+1}{a+k+1}\,
        {}_2F_1\left(1, \frac{k+1}a+1; \frac{k+1}a+2; -\dfrac1A\right)\,.
  \end{align}
My doubt and question is did I employ interchanging the infinite summation and integral on appropriate way? Is it possible or not?

I give you a hint to solve your integral analytical:
First you split the integral in two parts:
$$\int_0^{\infty } \frac{x^a}{(1+x)^2 \left(A+x^a\right)} \, \
dx=\int_0^{\infty } \frac{1}{(1+x)^2} \, dx-\int_0^{\infty } \
\frac{A}{(1+x)^2 \left(A+x^a\right)}\, dx$$
The first integral is trivial:
$$\int_0^{\infty } \frac{x^a}{(1+x)^2 \left(A+x^a\right)} \, dx=1-A \
\int_0^{\infty } \frac{1}{(1+x)^2 \left(A+x^a\right)} \, dx$$
Now you have to do two times partial integration to reduce the problem on an integral including the term $\text{log}[x+1]$:
$$\int_0^{\infty } \partial _x\left(-\frac{1}{(1+x) \left(A+x^a\right)}\right) \, dx=\int_0^{\infty } \frac{a x^{-1+a}}{(1+x) \
\left(A+x^a\right)^2}\, dx+\int_0^{\infty } \frac{1}{(1+x)^2 \left(A+x^a\right)} \, dx$$
$$\text{F6}(A,a)=1+A \left(-\frac{1}{A}+\int_0^{\infty } \frac{a \
x^{-1+a}}{(1+x) \left(A+x^a\right)^2} \, dx\right)$$
Take the integral on the right side to repeat the partial integration:
$$\int_0^{\infty } \partial _x\frac{\text{Log}[1+x] \left(a x^{-1+a}\right)}{\left(A+x^a\right)^2} \, dx=\int_0^{\infty } \frac{(-1+a) a \
x^{-2+a}\text{Log}[1+x]}{\left(A+x^a\right)^2} \, dx-\int_0^{\infty } \frac{2 a^2 x^{-2+2 a} \text{Log}[1+x]}{\left(A+x^a\right)^3} \, \
dx+\int_0^{\infty} \frac{a x^{-1+a}}{(1+x) \left(A+x^a\right)^2} \, dx$$
This leads to the integrals:
$$\text{F11}[(A,a)=\kappa [a,A] \int_0^{\infty } \frac{x^{2 (a-1)} \text{Log}[1+x]}{\left(A+x^a\right)^3} \, dx-\lambda [a,A] \
\int_0^{\infty } \frac{x^{-2+a}\text{Log}[1+x]}{\left(A+x^a\right)^2} \, dx$$
with the abbreviations:
$$\kappa (A,a)=2 a^2 A$$
and
$$\lambda (A,a)=A a (-1+a)$$
The first and second integral on the right hand side in a similar form are already solved by Sasha using the 
Plancherel theorem. 
For a similar derivation to [Sasha] we start with the introduction of k:
$$\text{F12}(k,A,a)=\int_0^{\infty } \frac{t^{a (k+1)} \text{Log}[1+t]}{\left(A+t^a\right)^{k+2} t^2} \, dt$$
to write 
$$\text{F13}(A,a)=\kappa (A,a) \text{F12}(1,A,a)-\lambda (A,a) \text{F12}(0,A,a)$$
Now we only have to calculate the $F12(k,A,a)$. For the calculation of $F12(k,A,a)$ we use the calculation of [Sasha] and take over the derivation of the $G_{1}\left( t\right)$, the only difference between our calculation and [Sacha] is the function $G_{2}\left( t\right)$ 
Mathematica has some problems to get the Inverse Mellin - Transform for general values of A and a. So we calculate for some different values of a and A:
$$\text{G21}(k,A,a)=\int_0^{\infty } \frac{t^{a (k+1)} \
t^{s-2}}{\left(A+t^a\right)^{k+2}} \, dt$$
$$\text{H21}(k,A,a)=\frac{\text{Gamma}[1-s] \text{Gamma}[s]^2 \text{G21}\left(k,A,\frac{1}{\alpha }\right)}{\text{Gamma}[s+1]}$$
$$\text{h21}(k,A,a)=\text{InverseMellinTransform}[\text{H21}(A,\lambda ),s,t]$$
$$\text{h21}(3,2)=\frac{2^{1+k} \text{MeijerG}\left[\left\{\left\{-1,-\frac{1}{2},0\right\},\{1\}\right\},\left\{\left\{0,0,\frac{1}{2} (-1+k),\frac{k}{2}\right\},\{\}\right\},\frac{t}{9}\right]}{27\pi  \text{Gamma}[2+k]}$$
other values are calculated in the same way. Final we get:
$$F14\left( A,k,a\right) =\frac{1}{a~A^{\frac{1}{a}+1}\left( k+1\right) !}
H_{3,3}^{3,2}\left( \frac{1}{A^{\frac{1}{a}}}\left\vert 
\begin{array}{c}
\left( -\frac{1}{a},\frac{1}{a}\right) ,\left( 0,1\right) ,\left( 1,1\right) 
\\ 
\left( 0,1\right) ,\left( 0,1\right) ,\left( k-\left( \frac{1}{a}-1\right) ,%
\frac{1}{a}\right) 
\end{array}\right. \right) $$
For $a=\frac{1}{2}$ we get:
$$F14\left( A,k,\frac{1}{2}\right) =\frac{2^{k+1}}{A^{3}\pi \left( k+1\right) !}%
G_{4,4}^{4,3}\left( \frac{1}{A^{2}}\left\vert 
\begin{array}{c}
-1,-\frac{1}{2},0,1 \\ 
0,0,\frac{k-1}{2},\frac{k}{2}%
\end{array}%
\right. \right) $$
Finally we get:
$$\int_0^{\infty } \frac{x^a}{(1+x)^2 \left(A+x^a\right)} \, \
dx=\kappa (a,A) \text{F14}(A,1,a)-\lambda (a,A) \text{F14}(A,0,a)$$
So the general solution of integral can be expressed in the form of two H-Fox-Functions, which are discussed in detail in 
Mathai
