Integration! Can you do this?

$I(a)=\displaystyle\int_{0}^{1} \frac{tan^{-1}ax}{x\sqrt{1-x^{2}}} dx$

By using Leibniz's formula,

$I'(a)=\displaystyle\int_{0}^{1} \frac{\partial}{\partial a} \frac{tan^{-1}ax}{x\sqrt{1-x^{2}}} dx$ $=\displaystyle\int_{0}^{1} \frac{x}{1+a^{2}x^{2}}×\frac{1}{x\sqrt{1-x^{2}}} dx$

Substituting $x=\frac{1}{t}$

$I'(a)=\displaystyle\int_{1}^{∞} \frac{t}{(a^{2}+t^{2})\sqrt{t^{2}-1}} dt$

Substituting $t^{2}-1=p^{2}$

$I'(a)=\displaystyle\int_{0}^{∞} \frac{p}{(a^{2}+p^{2}+1)|p|} dp$

$=\frac{1}{\sqrt{1+a^{2}}} \bigg [ tan^{-1} \frac{x}{\sqrt{1+a^{2}}} \bigg ]_{0}^{∞}$

$\Rightarrow\frac{dI}{da}=\frac{\pi}{2\sqrt{1+a^{2}}}$

$\Rightarrow I=\displaystyle\int \frac{\pi}{2\sqrt{1+a^{2}}} da$ $=\frac{\pi}{2} ln (a+\sqrt{a^{2}+1})+c$

By substituting a=0 in original equation, one may get c=0.

Hence, $I(a)=\frac{\pi}{2} ln (a+\sqrt{a^{2}+1})$

$I(3)=\frac{\pi}{2} ln (3+\sqrt{10})$

$p=2,\ q=3,\ r=10,\ s=0$

So $p+q+r+s=\boxed{15}$