Find Ratio of Integrals $I:J$ Given:
$$I=\int_{0}^{1}\frac{x^{\frac{5}{2}}(1-x)^{\frac{7}{2}}\:dx}{12}$$
and
$$J=\int_{0}^{1}\frac{x^{\frac{5}{2}}(1-x)^{\frac{7}{2}}\:dx}{(x+3)^8}$$
Find Value of $\frac{I}{J}$
My attempt:The Integral $I$ is easy to evaluate using Beta Function.
So i was trying to Manipulate $J$ to convert it to $I$ as follows:
We can write $J$ as:
$$J=\int_{0}^{1}\frac{x^{6}\left ( \frac{1}{x} -1\right )^\frac{7}{2}}{x^{8}\left ( 1+\frac{3}{x} \right )^8}$$
Now put $\frac{1}{x}=t$ we get:
$$J=\int_{1}^{\infty}\frac{(t-1)^{\frac{7}{2}}}{(1+3t)^8}$$
Using integration by Parts taking $u=(t-1)^{3.5}$ and $v=\frac{1}{(1+3t)^8}$ we get:
$$J=\frac{1}{6}\times \int_{1}^{\infty}\frac{(t-1)^{\frac{5}{2}}}{(1+3t)^7}$$
Repeating Parts again and again:
$$J=\frac{1}{432}\times \int_{1}^{\infty}\frac{\sqrt{t-1}}{(1+3t)^5}\:dt$$
Any way to proceed from here?
 A: Well, substitute $\sqrt{t-1}=z$
then, 
$\dfrac{1}{2}\cdot\dfrac{1}{\sqrt{t-1}}dt=dz$
Now multiply the J by $\dfrac{\sqrt{t-1}}{\sqrt{t-1}}$
$J=\dfrac{1}{216}\cdot\int^{\infty}_{1}\dfrac{\sqrt{t-1}}{(1+3t)^{5}}\cdot\dfrac{\sqrt{t-1}}{2\sqrt{t-1}}dt$
$J=\dfrac{1}{216}\cdot\int^{\infty}_{1}\dfrac{t-1}{(1+3t)^{5}}\cdot\dfrac{1}{2\sqrt{t-1}}dt$
Now, substitute $t=z^{2}+1$ and $\dfrac{1}{2\sqrt{t-1}}dt=dz$
$J=\dfrac{1}{216}\cdot\int^{\infty}_{0}\dfrac{z^{2}}{(3z^{2}+4)}dz$
Now use Integration by parts.
Hint:
$\dfrac{d}{dx}[x-tan^{-1}(x)]=\dfrac{x^{2}}{x^{2}+1}$
A: As you noted, $$I=\frac 1 {12}B\left(\frac 7 2, \frac 9  2\right)$$
Also, if $|z|<1$ and $\alpha \in \mathbb R$
$$\frac 1 {(1+z)^\alpha}  = \sum_{n\geq  0}\frac{(-\alpha)(-\alpha-1)...(-\alpha-n+1)}{n!}z^n\tag{1}$$
So for $\alpha =8$
$$\begin{split}
J &= \frac 1 {3^8}\int_0^1\frac{x^{\frac{5}{2}}(1-x)^{\frac{7}{2}}\:dx}{\left(1+\frac x 3\right)^8}\\
&=\frac 1 {3^8}\sum_{n\geq 0}\frac{(-8)(-9)...(-8-n+1)}{n!}\frac 1 {3^n}\int_0^1x^{\frac{5+n}{2}}(1-x)^{\frac{7}{2}}dx\\
&= \frac 1 {3^8}\sum_{n\geq 0}\frac{(-8)(-9)...(-8-n+1)}{n!}\frac 1 {3^n}B \left(\frac 7 2 +n,\frac 9 2\right)
\end{split}$$
Now, because $$B(a+1, b) = B(a,b)\frac a {a+b}$$
we also have
$$B(a+n, b)=B(a,b)\cdot\frac{a}{a+b}\cdot\frac{a+1}{a+b+1}...\frac{a+n-1}{a+b+n-1}$$
Also note that $\frac 7 2 +\frac 9 2 = 8$. Therefore 
$$B\left(\frac 7 2+n, \frac 9 2 \right)=B\left(\frac 7 2,\frac 9 2\right)\cdot\frac{\frac 7 2}{8}\cdot\frac{\frac 7 2+1}{9}...\frac{\frac 7 2+n-1}{8+n-1}$$
We now have
$$\begin{split}
J &=B\left(\frac 7 2,\frac 9  2\right)\frac 1 {3^8}\sum_{n\geq 0}\frac{(-\frac 7 2)(-\frac 7 2 -  1)...(-\frac 7 2-n+1)}{n!}\frac {1} {3^n}\\
&= B\left(\frac 7 2,\frac 9  2\right)\frac 1 {3^8}\frac 1 {(1+\frac 1 3)^{\frac 7 2}} \,\,\,\,\text{  (using (1))}\\
&=B\left(\frac 7 2,\frac 9  2\right)\frac 1 {3^{\frac 9 2}4^{\frac 7 2}}
\end{split}$$
Conclusion:
$$\boxed{\frac I J = \frac {3^{\frac 9 2}4^{\frac 7 2}}{12}=3^{\frac 7 2}4^{\frac 5 2}}$$
