If $I_1=\underset{0}{\overset{1}{\int }}\frac{x^{\raisebox{1ex}{$7$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{(1-x)}^{\raisebox{1ex}{$9$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}{30}dx$ and $I_2=\underset{0}{\overset{1}{\int }}\frac{x^{\raisebox{1ex}{$7$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{(1-x)}^{\raisebox{1ex}{$9$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}{{(x+5)}^{10}}dx$ and $\frac{I_1}{I_2}=5a^3\sqrt{a}$ where $a\in N$, Then the value of a is

 $I_2={{\int }^{\text{}}}_0^1{\left(\frac{x}{5+x}\right)}^{\frac{7}{2}}\cdot {\left(\frac{1-x}{5+x}\right)}^{\frac{9}{2}}\cdot \frac{dx}{{(5+x)}^2}$
Put  $\frac{x}{5+x}=t$
$\frac{5}{{(5+x)}^2}dx=dt\Rightarrow \frac{dx}{{(5+x)}^2}=\frac{1}{5}dt$
$\therefore I_2={{\int }^{\text{}}}_0^{\frac{1}{6}}{\left(t\right)}^{\frac{7}{2}}{(1-6t)}^{\frac{9}{2}}\frac{dt}{{\left(5\right)}^{11/2}};$ Now   Put $6t=\mu$ 
and simplify we get $I_2=\frac{1}{5^{9/2}\times 6^{7/2}}{I}_1$  we Conclude  $a=30$