<s>[INST] 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?
 [/INST] 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:
$\dfrac{d}{dx}[x-tan^{-1}(x)]=\dfrac{x^{2}}{x^{2}+1}$
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