यदि int_{0}^{sqrt{3}} frac{15 x^{3}}{sqrt{1+x | vedclass

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(A) माना $1 + x^{2} = t^{2}$. तब $2x dx = 2t dt$,अतः $x dx = t dt$.जब $x = 0$,तब $t = 1$. जब $x = \sqrt{3}$,तब $t = 2$.समाकलन $\int_{1}^{2} \frac{15(t

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$10$

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(A) माना $1 + x^{2} = t^{2}$. तब $2x dx = 2t dt$,अतः $x dx = t dt$.जब $x = 0$,तब $t = 1$. जब $x = \sqrt{3}$,तब $t = 2$.समाकलन $\int_{1}^{2} \frac{15(t^{2}-1) t dt}{\sqrt{t^{2} + t^{3}}} = 15 \int_{1}^{2} \frac{t(t^{2}-1)}{t \sqrt{1+t}} dt = 15 \int_{1}^{2} \frac{t^{2}-1}{\sqrt{1+t}} dt$ हो जाता है।माना $1 + t = u^{2}$,अतः $t = u^{2} - 1$ और $dt = 2u du$.जब $t = 1$,तब $u = \sqrt{2}$. जब $t = 2$,तब $u = \sqrt{3}$.समाकलन $15 \int_{\sqrt{2}}^{\sqrt{3}} \frac{(u^{2}-1)^{2}-1}{u} (2u du) = 30 \int_{\sqrt{2}}^{\sqrt{3}} (u^{4} - 2u^{2}) du$ हो जाता है।समाकलन का मान: $30 \left[ \frac{u^{5}}{5} - \frac{2u^{3}}{3} \right]_{\sqrt{2}}^{\sqrt{3}} = 30 \left[ \left( \frac{9\sqrt{3}}{5} - \frac{6\sqrt{3}}{3} \right) - \left( \frac{4\sqrt{2}}{5} - \frac{4\sqrt{2}}{3} \right) \right]$.$= 30 \left[ \left( \frac{9\sqrt{3} - 10\sqrt{3}}{5} \right) - \left( \frac{12\sqrt{2} - 20\sqrt{2}}{15} \right) \right] = 30 \left[ -\frac{\sqrt{3}}{5} + \frac{8\sqrt{2}}{15} \right] = -6\sqrt{3} + 16\sqrt{2}$.$\alpha \sqrt{2} + \beta \sqrt{3}$ से तुलना करने पर,$\alpha = 16$ और $\beta = -6$ प्राप्त होता है।अतः,$\alpha + \beta = 16 - 6 = 10$.

$x$	$1$	$2$	$3$	$4$
$y$	$0.7111$	$0.7222$	$0.7333$	$0.7444$

यदि $\int_{0}^{\sqrt{3}} \frac{15 x^{3}}{\sqrt{1+x^{2}+\sqrt{(1+x^{2})^{3}}}} dx = \alpha \sqrt{2} + \beta \sqrt{3}$,जहाँ $\alpha, \beta$ पूर्णांक हैं,तो $\alpha + \beta$ का मान ज्ञात कीजिए।

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