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
    $13$

Answer

Correct option: A.
$10$
a
Put $1+ x ^{2}= t ^{2}$

$2 x dx =2 t dt$

$X dx = t d t$

$\therefore \int_{1}^{2} \frac{15\left( t ^{2}-1\right) t dt }{\sqrt{ t ^{2}+ t ^{3}}}$

$15 \int_{1}^{2} \frac{ t \left( t ^{2}-1\right)}{ t \sqrt{1+ t }} dt$

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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