If $$\int\limits_{0}^{\sqrt{3}} \frac{15 x^{3}}{\sqrt{1+x^{2}+\sqrt{\left(1+x^{2}\right)^{3}}}} \mathrm{~d} x=\alpha \sqrt{2}+\beta \sqrt{3}$$, where $\alpha, \beta$ are integers, then $\alpha+\beta$ is equal to __________.

<p>Put $x = \tan \theta \Rightarrow dx = {\sec ^2}\theta \,d\theta$</p> <p>$$ \Rightarrow I = \int\limits_0^{{\pi \over 3}} {{{15{{\tan }^3}\theta \,.\,{{\sec }^2}\theta \,d\theta } \over {\sqrt {1 + {{\tan }^2}\theta + \sqrt {{{\sec }^6}\theta } } }}} $$</p> <p>$$ \Rightarrow I = \int\limits_0^{{\pi \over 3}} {{{15{{\tan }^2}\theta {{\sec }^2}\theta \,d\theta } \over {\sec \theta \sqrt {1 + \sec \theta } }}} $$</p> <p>$$ \Rightarrow I = \int\limits_0^{{\pi \over 3}} {{{15({{\sec }^2}\theta - 1)\sec \theta \tan \theta \,d\theta } \over {\left( {\sqrt {1 + \sec \theta } } \right)}}} $$</p> <p>Now put $1 + \sec \theta = {t^2}$</p> <p>$\Rightarrow \sec \theta \tan \theta \,d\theta = 2tdt$</p> <p>$$ \Rightarrow I = \int\limits_{\sqrt 2 }^{\sqrt 3 } {{{15\left( {{{({t^2} - 1)}^2} - 1} \right)2t\,dt} \over t}} $$</p> <p>$$ \Rightarrow I = 30\int\limits_{\sqrt 2 }^{\sqrt 3 } {({t^4} - 2{t^2} + 1 - 1)\,dt} $$</p> <p>$\Rightarrow I = 30\int\limits_{\sqrt 2 }^{\sqrt 3 } {({t^4} - 2{t^2})\,dt}$</p> <p>$$ \Rightarrow I = \left. {30\left( {{{{t^5}} \over 5} - {{2{t^3}} \over 3}} \right)} \right|_{\sqrt 2 }^{\sqrt 3 }$$</p> <p>$$ = 30\left[ {\left( {{9 \over 5}\sqrt 3 - 2\sqrt 3 } \right) - \left( {{{4\sqrt 2 } \over 5} - {{4\sqrt 2 } \over 3}} \right)} \right]$$</p> <p>$$ = \left( {54\sqrt 3 - 60\sqrt 3 } \right) - \left( {24\sqrt 2 - 40\sqrt 2 } \right)$$</p> <p>$= 16\sqrt 2 - 6\sqrt 3$</p> <p>$\therefore$ $\alpha = 16$ and $\beta = - 6$</p> <p>$\alpha + \beta = 10.$</p>