How to integrate this?Is it a standard integral? Is there a general theory for this integral? I read it at a textbook(not an english one) and the author just throws the answer.
It seems like it is a standard integral but i can't find any info on it.
$$\int_0^{\infty} \frac{x^{2p-1}}{(ax^2+b)^{p+q}}=\frac{B(p,q)}{2a^pb^q}$$ 
 A: $$
\int_0^{\infty} \frac{x^{2p-1} dx}{(ax^2+b)^{p+q}}=\frac{1}{b^{p+q}}\int_0^{\infty} \frac{x^{2p-1}dx}{((a/b)x^2+1)^{p+q}}\ ,
$$
then change variables $(a/b)x^2=t\Rightarrow 2(a/b)xdx=dt$ to obtain
$$
\frac{1}{b^{p+q}}\frac{b}{2a}(b/a)^{p-1}\int_0^{\infty} \frac{t^{p-1}dt}{(t+1)^{p+q}}\ ,
$$
and then use the identity
$$
\mathrm{B}(x,y)=\int_0^{\infty}dt\frac{t^{x-1}}{(1+t)^{x+y}}\ ,
$$
where $\mathrm{B}(x,y)$ is Euler's Beta function.
A: First pull the constant $a$ and $b$ away by a scaling of the variable, and get rid of the square with $ax^2/b\to t$.
To a constant factor, you now have
$$\int_0^\infty\frac{t^{p-1}}{(t+1)^{p+q}}dt.$$
Now you convert to the classical Beta form by $u:=t/(t+1)$,
$$\int_0^1u^{p-1}(1-u)^{q-1}du.$$

The value of this integral can be obtained using the Gamma function. The standard proof relies on the decomposition of a particular double integral as a product of integrals. It bears some resemblance to the convolution theorem of the Laplace transform.
A: This integral of yours is evaluated using Beta function. https://en.wikipedia.org/wiki/Beta_function
