The value of the integral $I = \int\limits_0^1\frac{1}{\sqrt{1+3\sqrt{x}}}$ is

16/27

The correct answer is Option (1) → 16/27

Given

$I=\int_{0}^{1}\frac{1}{\sqrt{1+3\sqrt{x}}}\,dx$

Put $t=\sqrt{x}$

$x=t^2,\;dx=2t\,dt$

When $x=0,\;t=0$ and when $x=1,\;t=1$

$I=\int_{0}^{1}\frac{2t}{\sqrt{1+3t}}\,dt$

Put $u=1+3t$

$du=3dt,\;t=\frac{u-1}{3}$

$dt=\frac{du}{3}$

$I=\int_{u=1}^{4}\frac{2\frac{u-1}{3}}{\sqrt{u}}\cdot\frac{du}{3}$

$=\frac{2}{9}\int_{1}^{4}\frac{u-1}{\sqrt{u}}\,du$

$=\frac{2}{9}\int_{1}^{4}\left(u^{\frac12}-u^{-\frac12}\right)du$

$=\frac{2}{9}\left[\frac{2}{3}u^{\frac32}-2u^{\frac12}\right]_{1}^{4}$

$=\frac{2}{9}\left[\left(\frac{2}{3}\cdot8-2\cdot2\right)-\left(\frac{2}{3}\cdot1-2\cdot1\right)\right]$

$=\frac{2}{9}\left[\left(\frac{16}{3}-4\right)-\left(\frac{2}{3}-2\right)\right]$

$=\frac{2}{9}\left[\frac{4}{3}+\frac{4}{3}\right]$

$=\frac{2}{9}\cdot\frac{8}{3}$

$=\frac{16}{27}$

The value of the integral is $\frac{16}{27}$.