\(u = tan^{-1} (x + \sqrt {1 + x^2})\)

\(v = \frac {tan^{-1}x}{1 + \sqrt{1 + x^2}}\)

\(\frac{d u}{d x} = \frac{1}{1 + (x + \sqrt{1 + x^2})^2} \;\frac {d}{dx} (x + \sqrt{1+ x^2})\)

\(= \frac{1}{2(x^2 + 1) + 2x\sqrt{1 + x^2}} \times \left(1 + \frac{2x}{2\sqrt{1 + x^2}}\right)\)

\(= \frac 1{2\sqrt{x^2 + 1} .(\sqrt{x^2 + 1}+x)} \times \frac {x + \sqrt{1 +x^2}}{\sqrt {1 +x^2}}\)

\(=\frac 1{2(1 +x^2)}\)

\(\frac{d v}{d x} = \cfrac{(1 + \sqrt {1 +x^2})\times \frac 1{1+x^2} - tan^{-1}x \times\frac{2x}{2\sqrt{1 + x^2}}}{(1+ \sqrt{1 +x^2})^2}\)

\(=\cfrac {\frac{1 + \sqrt{1 +x^2}}{1 + x^2} - \frac{xtan^{-1}x}{\sqrt{1 + x^2}}}{( 1 + \sqrt{1 + x^2})^2}\)

\(= \frac{1 + \sqrt {1 + x^2} - x\sqrt{1 + x^2}tan^{-1}x}{( 1 + x^2) ( 1+ \sqrt{ 1 +x^2})^2}\)

\(\cfrac{d\,tan^{-1} (x + \sqrt{ 1+ x^2})}{d\left(\frac{tan^{-1}x}{1 + \sqrt{1 + x^2}}\right)} = \frac {du}{dv} = \cfrac{\frac{du}{dx}}{\frac{d v}{d x}}\)

\(= \cfrac{\frac1{2( 1 + x^2)}}{\frac{1+ \sqrt{1+ x^2} (1- x\,tan^{-1}x)}{(1+x^2) (1+ \sqrt{1+ x^2})^2}} \)

\(= \frac{(1 +\sqrt{1 +x^2})^2}{2(1 + \sqrt{ +x^2}( 1 - x \, tan^{-1}x))}\)