This will come in handy!

Let $u = \sqrt{\frac{4x^2+1}{x^2-1}}.$ Then $u^2-4 = \frac5{x^2-1},$ so $x^2 = \frac{u^2+1}{u^2-4}$ and $2x \, dx = \frac{-10u \, du}{(u^2-4)^2}.$

Hence $\frac{dx}{x} = (x \, dx) \frac1{x^2} = \frac{-5u \, du}{(u^2-4)^2} \frac{u^2-4}{u^2+1} = -\frac{5u \, du}{(u^2+1)(u^2-4)}.$

Then the integral becomes $\begin{aligned} \int_3^\infty \frac{5u^2}{(u^2+1)(u^2-4)} \, du &= \int_3^\infty \left( \frac1{u^2+1} + \frac4{u^2-4} \right) \, du \\ &= \left( \text{arctan}(u) + \ln\left( \frac{u-2}{u+2} \right) \right) \bigg|_3^\infty \\ &= \pi/2 - \left( \text{arctan}(3) + \ln\left( \frac{3-2}{3+2} \right) \right) \\ &= \pi/2 - \text{arctan}(3) - \ln(1/5) = \frac{\pi}2 - \text{arctan}(3) + \ln(5). \end{aligned}$ So the answer is $2+5+3 = \fbox{10}.$

The successive substitutions $u^2 = 4x^2 + 1$ , $u = \sqrt{5}\sec v$ , $2w = \sin v$ give $\begin{aligned} \int \sqrt{\frac{4x^2+1}{x^2-1}}\,\frac{dx}{x} & = \; 2\int \frac{u^2\,du}{(u^2-1)\sqrt{u^2-5}} \\ & = \; 2\int \frac{du}{\sqrt{u^2 - 5}} + 2\int \frac{du}{(u^2-1)\sqrt{u^2-5}} \; = \; 2\cosh^{-1}\big(\tfrac{u}{\sqrt{5}}\big) + 2\int \frac{\cos v\,dv}{4 + \sin^2v} \\ & = \; 2\cosh^{-1}\left(\sqrt{\tfrac{4x^2+1}{5}}\right) + \int \frac{dw}{1 + w^2} \; = \; 2\cosh^{-1}\left(\sqrt{\tfrac{4x^2+1}{5}}\right) + \tan^{-1}w + k \\ & = \; 2\cosh^{-1}\left(\sqrt{\tfrac{4x^2+1}{5}}\right) + \tan^{-1}\left(\sqrt{\tfrac{x^2-1}{4x^2+1}}\right) + k \end{aligned}$ for an arbitrary constant $k$ , and hence $\begin{aligned} \int_1^{\sqrt{2}} \sqrt{\frac{4x^2+1}{x^2-1}}\,\frac{dx}{x} & = \; \left[ 2\cosh^{-1}\left(\sqrt{\tfrac{4x^2+1}{5}}\right) + \tan^{-1}\left(\sqrt{\tfrac{x^2-1}{4x^2+1}}\right) \right]_1^{\sqrt{2}} \\ & = \; 2\cosh^{-1}\tfrac{3}{\sqrt{5}} + \tan^{-1}\tfrac13 \; = \; \tfrac12\pi + \ln5 - \tan^{-1}3 \end{aligned}$

making the answer $2 + 5 + 3 = \boxed{10}$ .

Substituting $x=\cosh(t)$ .

$\int_a^{b}{\frac{4\cosh^2(t)+1}{\sinh(t)}\frac{\sinh(t)}{\cosh(t)}}dt= \int_a^{b}{\frac{4\cosh^2(t)+1}{\cosh(t)}}dt$ where $a=0$ and $b=\cosh^{-1}(\sqrt2)$ . Wolfram alpha helped me on this, resulting in

$2\sinh^{-1}(\frac{2\sinh(t)}{\sqrt(5)}) + \tan^{-1}(\frac{\sinh(t)}{\sqrt{2\cosh(2t)+3}})$ which evaluates to 0 for the lower bound $a=0$ , so that we just need the value at b.

Since b is positive, $\sinh(b)= \sqrt{\cosh^2(b)-1}=2-1=1$

Also, $\cosh(2t)=2\cosh^2(t)-1$ , so our answer simplifies into

$2\sinh^{-1}(\frac{2}{\sqrt{5}}) + \tan^{-1}(\frac{1}{\sqrt{4\cosh^2(b)+1}})$

Filling in $b=\cosh^{-1}(\sqrt(2))$ shows that $\sqrt{4\cosh^2(b)+1}=3$ , and

Solving $\sinh(y)= \frac{2}{\sqrt{5}} \implies (e^{y})^2-\frac{4}{\sqrt{5}}e^y-1=0 \implies e^y=\sqrt5 \implies y=\ln(\sqrt5)$ (discarding the imaginary solution) $\implies 2\sinh^{-1}(\frac{2}{\sqrt{5}})=2\ln(\sqrt{5})=\ln(5)$

Finally $tan^{-1}(\frac{1}{x})=\frac{\pi}{2}-tan^{-1}(x)$ so we arrive at

$ln(5) + \frac{\pi}{2}-\tan^{-1}(3)$ The requested answer is $2+5+3=10$ .