Let $\alpha 0$. If ${\int }_{0}^{\alpha }\frac{x}{\sqrt{x+\alpha }-\sqrt{x}}dx=\frac{16+20\sqrt{2}}{15}$, then $\alpha$ is equal to: [JEE Main 2023, 31 Jan (Shift 2)]

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Let $\alpha >0$. If ${\int }_{0}^{\alpha }\frac{x}{\sqrt{x+\alpha }-\sqrt{x}}dx=\frac{16+20\sqrt{2}}{15}$, then $\alpha$ is equal to: [JEE Main 2023, 31 Jan (Shift 2)]

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Let \(\alpha >0\). If \({\int }_{0}^{\alpha }\frac{x}{\sqrt{x+\alpha }-\sqrt{x}}dx=\frac{16+20\sqrt{2}}{15}\), then \(\a…

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