Let α > 0 . If ∫ _ 0 ^ α x x + α - x dx = 16 + 20 2 15 then α is...

Given,

${∫}_{0}^{\alpha}\dfrac{x}{\sqrt{x + \alpha} - \sqrt{x}}dx = \dfrac{16 + 20\sqrt{2}}{15} . . . . ( 1 )$

Rationalise the denominator in L.H.S we get,

${∫}_{0}^{\alpha}\dfrac{x ( \sqrt{\alpha + x} + \sqrt{x} )}{\alpha}dx = \dfrac{1}{\alpha}\left[{∫}_{0}^{\alpha}x\sqrt{\alpha + x}dx + {∫}_{0}^{\alpha}{x}^{3 / 2}dx\right]$

Put $\alpha + x = {t}^{2} ⇒ dx = 2tdt$

Limit when $x = 0 \to t = \sqrt{\alpha}, x = \alpha \to t = \sqrt{2\alpha}$

$\dfrac{1}{\alpha}{∫}_{\sqrt{\alpha}}^{\sqrt{2\alpha}}\left({t}^{2} - \alpha\right)2{t}^{2}dt + \dfrac{2{\alpha}^{3 / 2}}{5}$

$= \dfrac{1}{\alpha}{\left[\dfrac{2{t}^{5}}{5} - \dfrac{2\alpha{t}^{3}}{3}\right]}_{\sqrt{\alpha}}^{\sqrt{2\alpha}} + \dfrac{2{\alpha}^{3 / 2}}{5}$

$= \dfrac{4\alpha\sqrt{2\alpha} + 10\alpha\sqrt{\alpha}}{15}$

From equation \left(1\right)$</p><p>${∫}_{0}^{\alpha}\dfrac{x}{\sqrt{x + \alpha} - \sqrt{x}}dx = \dfrac{16 + 20\sqrt{2}}{15}$</p><p>$⇒ \dfrac{4\alpha\sqrt{2\alpha} + 10\alpha\sqrt{\alpha}}{15} = \dfrac{16 + 20\sqrt{2}}{15}$</p><p>On comparing both sides we get,</p><p>$\alpha = 2$