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  4. Let α>0. If ∫ limits0α (x/√x+α-√x) d x=(16+20 √2/15), then α is equal to :

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Q. Let $\alpha>0$. If $\int\limits_0^\alpha \frac{x}{\sqrt{x+\alpha}-\sqrt{x}} d x=\frac{16+20 \sqrt{2}}{15}$, then $\alpha$ is equal to :

JEE MainJEE Main 2023Integrals

Solution:

After rationalising
$ \int\limits_0^\alpha \frac{ x }{\alpha}(\sqrt{ x +\alpha}+\sqrt{ x }) $
$ \int\limits_0^\alpha \frac{1}{\alpha}\left[( x +\alpha)^{3 / 2}-\alpha( x +\alpha)^{1 / 2}+ x ^{3 / 2}\right]$
$ \left.\frac{1}{\alpha}\left[\frac{2}{5}( x +\alpha)^{5 / 2}-\alpha \frac{2}{3}( x +\alpha)^{3 / 2}+\frac{2}{5} x ^{5 / 2}\right]\right|_0 ^\alpha$
$ =\frac{1}{\alpha}\left(\frac{5}{2}(2 \alpha)^{5 / 2}-\frac{2 \alpha}{3}(2 \alpha)^{3 / 2}+\frac{2}{5} \alpha^{5 / 2}-\frac{2}{5} \alpha^{5 / 2}+\frac{2}{3} \alpha^{5 / 2}\right.$
$ =\frac{1}{\alpha}\left(\frac{2^{7 / 2} \alpha^{5 / 2}}{5} \frac{2^{5 / 2} \alpha^{5 / 2}}{3}+\frac{2}{3} \alpha^{5 / 2}\right) $
$ =\alpha^{3 / 2}\left(\frac{2^{7 / 2}}{5}-\frac{2^{5 / 2}}{3}+\frac{2}{3}\right) $
$ =\frac{\alpha^{3 / 2}}{15}(24 \sqrt{2}-20 \sqrt{2}+10)=\frac{\alpha^{3 / 2}}{15}(4 \sqrt{2}+10)$
Now,
$ \frac{\alpha^{3 / 2}}{15}(4 \sqrt{2}+10)=\frac{16+20 \sqrt{2}}{15}$
$ \Rightarrow \alpha=2$