The value of the integral I = ∫_0^1 1/(√(1+3√x)) dx is

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

Let \(u = 1 + 3\sqrt{x}\). This substitution removes both roots.

    \( \displaystyle \sqrt{x} = \frac{u - 1}{3},\quad x = \frac{(u - 1)^2}{9} \)

    \( \displaystyle dx = \frac{2(u - 1)}{9}\,du \)

Changing the limits: x = 0 gives u = 1, and x = 1 gives u = 4.

    \( \displaystyle I = \int_{1}^{4} \frac{1}{\sqrt{u}} \cdot \frac{2(u - 1)}{9}\,du \)

    \( \displaystyle I = \frac{2}{9}\int_{1}^{4} \left(u^{1/2} - u^{-1/2}\right) du \)

    \( \displaystyle I = \frac{2}{9}\left[\frac{2}{3}u^{3/2} - 2u^{1/2}\right]_{1}^{4} \)

At u = 4 and at u = 1:

    \( \displaystyle \frac{2}{3}(8) - 2(2) = \frac{16}{3} - 4 = \frac{4}{3} \)

    \( \displaystyle \frac{2}{3}(1) - 2(1) = -\frac{4}{3} \)

    \( \displaystyle I = \frac{2}{9}\left(\frac{4}{3} + \frac{4}{3}\right) = \frac{2}{9} \cdot \frac{8}{3} = \frac{16}{27} \)