MCQ
If $\int_{0}^{\sqrt{3}} \frac{15 x^{3}}{\sqrt{1+x^{2}+\sqrt{\left(1+x^{2}\right)^{3}}}} d x=\alpha \sqrt{2}+\beta \sqrt{3}$, where $\alpha, \beta$ are integers, then $\alpha+\beta$ is equal to.
- ✓$10$
- B$11$
- C$12$
- D$13$
$2 x dx =2 t dt$
$X dx = t d t$
$\therefore \int_{1}^{2} \frac{15\left( t ^{2}-1\right) t dt }{\sqrt{ t ^{2}+ t ^{3}}}$
$15 \int_{1}^{2} \frac{ t \left( t ^{2}-1\right)}{ t \sqrt{1+ t }} dt$
Put $1+ t = u ^{2}$
$dt =2 u du$
$15 \int_{\sqrt{2}}^{\sqrt{3}} \frac{\left( u ^{2}-1\right)^{2}-1}{ u } \times 2 u d u$
$\sqrt{3}$
$30 \int_{\sqrt{2}}\left( u ^{4}-2 u ^{2}\right) du$
$30\left(\frac{ u ^{5}}{5}-\frac{2 u ^{3}}{3}\right)_{\sqrt{2}}^{\sqrt{3}}$
$30\left[\frac{1}{5}\left(\sqrt{3}^{5}-\sqrt{2}^{5}\right)-\frac{2}{3}\left(\sqrt{3}^{3}-\sqrt{2}^{3}\right)\right]$
$30\left[\frac{1}{5}(9 \sqrt{3}-4 \sqrt{2})-\frac{2}{3}(3 \sqrt{3}-2 \sqrt{2})\right]$
$30\left[-\frac{1}{5} \times \sqrt{3}+\frac{8}{15} \sqrt{2}\right]$
$-6 \sqrt{3}+16 \sqrt{2}=\alpha \sqrt{2}+\beta \sqrt{3}$
$\alpha=16, \beta=-6$
$\therefore \alpha+\beta=10$
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$\mathrm{f}(\mathrm{x})= \int_{0}^{x}[y] \,d y$
Where $[x]$ is the greatest integer less than or equal to $x$. Which of the following is true?