The value of the integral $$\int_0^1 \frac{\sqrt{x} dx}{(1+x)(1+3x)(3+x)}$$ is:

Step 1: Substitution

To eliminate the square root, let $$x = t^2$$. Then $$dx = 2t \, dt$$.

Changing the limits:

• When $$x = 0$$, $$t = 0$$.
• When $$x = 1$$, $$t = 1$$.

The integral becomes:

$$I = \int_{0}^{1} \frac{t \cdot 2t}{(1+t^2)(1+3t^2)(3+t^2)} dt = 2 \int_{0}^{1} \frac{t^2}{(1+t^2)(3t^2+1)(t^2+3)} dt$$

Step 2: Partial Fraction Decomposition

Let $$u = t^2$$. We decompose the integrand:

$$\frac{u}{(1+u)(1+3u)(3+u)} = \frac{A}{1+u} + \frac{B}{1+3u} + \frac{C}{3+u}$$

Multiplying through by the denominator:

$$u = A(1+3u)(3+u) + B(1+u)(3+u) + C(1+u)(1+3u)$$

Solving for constants:

• Set $$u = -1$$: $$-1 = A(1-3)(3-1) = -4A \implies A = \frac{1}{4}$$
• Set $$u = -1/3$$: $$-\frac{1}{3} = B(1-\frac{1}{3})(3-\frac{1}{3}) = B(\frac{2}{3})(\frac{8}{3}) = \frac{16}{9}B \implies B = -\frac{3}{16}$$
• Set $$u = -3$$: $$-3 = C(1-3)(1-9) = 16C \implies C = -\frac{3}{16}$$

Step 3: Integration

Substitute the partial fractions back into the integral:

$$I = 2 \int_{0}^{1} \left[ \frac{1}{4(1+t^2)} - \frac{3}{16(3t^2+1)} - \frac{3}{16(t^2+3)} \right] dt$$

$$I = 2 \left[ \frac{1}{4} \tan^{-1} t - \frac{3}{16\sqrt{3}} \tan^{-1} (\sqrt{3}t) - \frac{3}{16\sqrt{3}} \tan^{-1} \left(\frac{t}{\sqrt{3}}\right) \right]_{0}^{1}$$

Evaluate at the limits $$1$$ and $$0$$:

$$I = 2 \left[ \left( \frac{1}{4} \cdot \frac{\pi}{4} - \frac{\sqrt{3}}{16} \cdot \frac{\pi}{3} - \frac{\sqrt{3}}{16} \cdot \frac{\pi}{6} \right) - (0) \right]$$

$$I = \frac{\pi}{8} - \frac{\sqrt{3}}{8} \left( \frac{\pi}{3} + \frac{\pi}{6} \right)$$

$$I = \frac{\pi}{8} - \frac{\sqrt{3}}{8} \left( \frac{\pi}{2} \right) = \frac{\pi}{8} - \frac{\sqrt{3}\pi}{16}$$

Factoring out $$\frac{\pi}{8}$$:

$$I = \frac{\pi}{8} \left( 1 - \frac{\sqrt{3}}{2} \right)$$

Final Answer:

The correct option is C:

$$\frac{\pi}{8} \left( 1 - \frac{\sqrt{3}}{2} \right)$$