Question

If $\int\limits_0^{\sqrt{3}} \frac{15 x^3}{\sqrt{1+x^2+\sqrt{\left(1+x^2\right)^3}}} dx =\alpha \sqrt{2}+\beta \sqrt{3}$, where $\alpha, \beta$ are integers, then $\alpha+\beta$ is equal to

Answer 1

Put $1+x^2=t^2$
$2 dx =2 t dt$
$X dx = t dt$
$\therefore \int \limits_1^2 \frac{15\left( t ^2-1\right) tdt }{\sqrt{ t ^2+ t ^3}}$
$15 \int \limits_1^2 \frac{ t \left( t ^2-1\right)}{ t \sqrt{1+ t }} dt$
Put $1+ t = u ^2$
$dt =2 u du$
$15 \int \limits_{\sqrt{2}}^{\sqrt{3}} \frac{\left(u^2-1\right)^2-1}{u} \times 2 u d u$
$30 \int \limits_{\sqrt{2}}^{\sqrt{3}}\left(u^4-2 u^2\right) d u$
$30\left(\frac{ u ^5}{5}-\frac{2 u ^3}{3}\right)_{\sqrt{2}}^{\sqrt{3}}$
$30\left[\frac{1}{5}\left(\sqrt{3}^5-\sqrt{2}^5\right)-\frac{2}{3}\left(\sqrt{3}^3-\sqrt{2}^3\right)\right]$
$30\left[\frac{1}{5}(9 \sqrt{3}-4 \sqrt{2})-\frac{2}{3}(3 \sqrt{3}-2 \sqrt{2})\right]$
$30\left[-\frac{1}{5} \times \sqrt{3}+\frac{8}{15} \sqrt{2}\right]$
$-6 \sqrt{3}+16 \sqrt{2}=\alpha \sqrt{2}+\beta \sqrt{3}$
$\alpha=16, \beta=-6$
$\therefore \alpha+\beta=10$