Question related to beta and gamma function I'm trying to derive the following integral.
$$\int_0^\infty \frac{x^8(1-x^6)}{(1+x)^{24}} \, dx.$$
What transformations can I use?
 A: In general
$$\int^{\infty}_{0}\frac{x^{u-1} }{(1+bx)^{m+1}}dx=\frac{1}{b^{u}}\beta (u,m+1-u).$$
Proof:
We make the transformation $y=\frac{bx}{1+bx}$ so 
$$y-1 =\frac{bx}{1+bx} -1 = \frac{bx-1-bx}{1+bx} = \frac{-1}{1+bx}\Rightarrow $$ $$bx+1=\frac{1}{1-y} \Rightarrow (1-y)^{m+1}=\frac{1}{(1+bx)^{m+1}}.$$
We have,
 $$bx+1=\frac{1}{1-y} \Rightarrow bx= \frac{1}{1-y} -1 = \frac{1-1+y}{1-y}= y (1-y)^{-1}$$
so $x=\frac{y(1-y)^{-1}}{b}$ then it follows $dy=b^{-1}(1-y)^{-2}dx$, and the desired integral is
$$\frac{1}{b^{u}}\int^{1}_{0}y^{u-1}(1-y)^{m-u}dy=\frac{1}{b^{u}}\beta (u,m+1-u). $$
Example:
In your case we take $b=1$, $m+1=24$, and take $x^8 (1-x^6)=x^8-x^{14}$, and  the difference of the results
A: Here is a way to evaluate the integral without invoking the Beta function.
Writing the integral as
$$\int_0^\infty \frac{x^8 (1 - x^6)}{(1 + x)^{24}} \, dx = \int_0^1 \frac{x^8 (1 - x^6)}{(1 + x)^{24}} \, dx + \int_1^\infty \frac{x^8 (1 - x^6)}{(1 + x)^{24}} \, dx.$$
In the second of these integral, if a substitution of $x \mapsto 1/x$ is enforced we have
\begin{align*}
\int_0^\infty \frac{x^8 (1 - x^6)}{(1 + x)^{24}} \, dx &= \int_0^1 \frac{x^8 (1 - x^6)}{(1 + x)^{24}} \, dx + \int_0^1 \frac{\frac{1}{x^8} \left (1 - \frac{1}{x^6} \right )}{\left (1 + \frac{1}{x} \right )^{24}} \frac{dx}{x^2}\\
&= \int_0^1 \frac{x^8 (1 - x^6)}{(1 + x)^{24}} \, dx + \int_0^1 \frac{x^8 (x^6 - 1)}{(1 + x)^{24}} \, dx\\
&= \int_0^1 \frac{x^8 (1 - x^6)}{(1 + x)^{24}} \, dx - \int_0^1 \frac{x^8 (1 - x^6)}{(1 + x)^{24}} \, dx\\
&= 0.
\end{align*}
