Trigonotegral

$\frac { \sin { (\tan ^{ -1 }{ x } ) } }{ \cos { (\tan ^{ -1 }{ x } ) } } =\tan { (\tan ^{ -1 }{ x } )=x } ,\\ \frac { \sin { (\tan ^{ -1 }{ x } ) } }{ x } =\cos { (\tan ^{ -1 }{ x } ) } \\ 1-\cos ^{ 2 }{ (\tan ^{ -1 }{ x) } } ={ x }^{ 2 }\cos ^{ 2 }{ (\tan ^{ -1 }{ x) } } \\ \cos ^{ 2 }{ (\tan ^{ -1 }{ x) } } =\frac { 1 }{ 1+{ x }^{ 2 } } \Rightarrow \cos ^{ 2 }{ (\tan ^{ -1 }{ x) } } dx=\frac { dx }{ 1+{ x }^{ 2 } } \\ let\quad y=\quad \tan ^{ -1 }{ (x) } \quad ,\quad then\quad dy=\frac { dx }{ 1+{ x }^{ 2 } } =\cos ^{ 2 }{ (\tan ^{ -1 }{ x)\quad dx } } \\ \\ I=\int _{ 0 }^{ 1 }{ { \frac { { \sin { (\tan ^{ -1 }{ x } )\sqrt { \cos ^{ 2 }{ (\tan ^{ -1 }{ x } ) } +4{ x }^{ 2 } } } } }{ x } } } dx\\ \quad =\int _{ 0 }^{ 1 }{ \cos { (\tan ^{ -1 }{ x } )\sqrt { \cos ^{ 2 }{ (\tan ^{ -1 }{ x } )+4{ x }^{ 2 } } } } } dx\\ \quad =\int _{ \tan ^{ -1 }{ 0 } }^{ \tan ^{ -1 }{ 1 } }{ \frac { \sqrt { \cos ^{ 2 }{ (y)+4\tan ^{ 2 }{ (y) } } } }{ \cos { (y) } } } dy\\ \quad =\int _{ 0 }^{ \frac { \pi }{ 4 } }{ \sqrt { 1+4\tan ^{ 2 }{ (y)\sec ^{ 2 }{ (y) } } } dy } \\ \quad =\int _{ 0 }^{ \frac { \pi }{ 4 } }{ \sqrt { 1+4\tan ^{ 2 }{ (y)+4\tan ^{ 4 }{ (y) } } } dy } \\ \quad =\int _{ 0 }^{ \frac { \pi }{ 4 } }{ 1+2\tan ^{ 2 }{ (y)\quad dy } } \\ \quad =\int _{ 0 }^{ \frac { \pi }{ 4 } }{ 2\sec ^{ 2 }{ (y) } -1 } \quad dy\\ \quad =2\tan { (\frac { \pi }{ 4 } )-\frac { \pi }{ 4 } } \\ \quad =2-\frac { \pi }{ 4 } \\ \boxed{2+4=6}$

It is important to note that:

$\large \displaystyle \sin(\tan^{-1} x ) = \frac{x}{\sqrt{x^2+1}}$ and $\large \displaystyle \cos(\tan^{-1} x ) = \frac{1}{\sqrt{x^2+1}}$

So:

$\large \displaystyle \int_0^1 \frac{\sin(\tan^{-1} x ) \sqrt{\cos^2(\tan^{-1} x ) + 4x^2}}{x} dx$

$\large \displaystyle = \int_0^1 \frac{ \frac{x}{\sqrt{x^2+1}} \sqrt{ \frac{1}{x^2+1} + 4x^2 } }{x}$

$\large \displaystyle = \int_0^1 \frac{1}{x^2+1} \sqrt{1 + 4x^2 + 4x^4} dx$

$\large \displaystyle = \int_0^1 \frac{1}{x^2+1} \sqrt{(1 + 2x^2)^2} dx$

$\large \displaystyle = \int_0^1 \frac{2x^2+1}{x^2+1}dx$

$\large \displaystyle = \int_0^1 2 - \frac{1}{x^2+1}dx$

$\large \displaystyle = 2 - \arctan(x) \Bigg|_0^1$

$\color{#20A900} \boxed{\large \displaystyle = 2 -\frac{\pi}{4}}$

Then:

$\color{#3D99F6} A = 2, B = 4, \boxed{\large \displaystyle A+B = 6}$