The value of the integral $I = \int_0^1 \frac{1}{\sqrt{1+3\sqrt{x}}} dx$ is

Given:

$I = \int_0^1 \frac{1}{\sqrt{1+3\sqrt{x}}} dx$

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Let $u = \sqrt{x}$

Then $x = u^2$ and $dx = 2u \, du$

When $x = 0$: $u = 0$

When $x = 1$: $u = 1$

$I = \int_0^1 \frac{1}{\sqrt{1+3u}} \cdot 2u \, du$

$I = \int_0^1 \frac{2u}{\sqrt{1+3u}} du$

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Let $t = 1 + 3u$

Then $u = \frac{t-1}{3}$ and $du = \frac{1}{3} dt$

When $u = 0$: $t = 1$

When $u = 1$: $t = 4$

$I = \int_1^4 \frac{2 \cdot \frac{t-1}{3}}{\sqrt{t}} \cdot \frac{1}{3} dt$

$I = \int_1^4 \frac{2(t-1)}{9\sqrt{t}} dt$

$I = \frac{2}{9} \int_1^4 \frac{t-1}{\sqrt{t}} dt$

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$I = \frac{2}{9} \int_1^4 \left(\frac{t}{\sqrt{t}} - \frac{1}{\sqrt{t}}\right) dt$

$I = \frac{2}{9} \int_1^4 \left(t^{1/2} - t^{-1/2}\right) dt$

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$I = \frac{2}{9} \left[ \frac{t^{3/2}}{3/2} - \frac{t^{1/2}}{1/2} \right]_1^4$

$I = \frac{2}{9} \left[ \frac{2t^{3/2}}{3} - 2t^{1/2} \right]_1^4$

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At $t = 4$:

$\frac{2(4)^{3/2}}{3} - 2(4)^{1/2} = \frac{2(8)}{3} - 2(2)$

$= \frac{16}{3} - 4$

$= \frac{4}{3}$

At $t = 1$:

$\frac{2(1)^{3/2}}{3} - 2(1)^{1/2} = \frac{2}{3} - 2$

$= -\frac{4}{3}$

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

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

$I = \frac{16}{27}$

Therefore, the value of the integral is $\dfrac{16}{27}$.