Key Concepts and Formulas

• Substitution Rule for Definite Integrals: If $u = g(x)$ is a differentiable function whose range is an interval $I$, and $f$ is continuous on $I$, then $\int_a^b f(g(x))g'(x) dx = \int_{g(a)}^{g(b)} f(u) du$.
• Partial Fraction Decomposition: A technique used to express a rational function as a sum of simpler rational functions. For a rational function of the form $\frac{P(x)}{Q(x)}$, where $Q(x)$ can be factored, we can decompose it into simpler fractions. In this problem, we will treat $x$ as a variable and perform partial fraction decomposition on the expression involving $\sqrt{x}$.
• Standard Integrals: Knowledge of standard integral forms, particularly those involving inverse trigonometric functions like $\int \frac{1}{a^2+u^2} du = \frac{1}{a} \arctan\left(\frac{u}{a}\right) + C$.

Step-by-Step Solution

Step 1: Strategic Substitution To simplify the integrand, we observe the presence of $\sqrt{x}$. Let $u = \sqrt{x}$. Then $u^2 = x$, and differentiating both sides with respect to $u$, we get $2u \, du = dx$. We also need to change the limits of integration. When $x = 0$, $u = \sqrt{0} = 0$. When $x = 1$, $u = \sqrt{1} = 1$. Substituting these into the integral, we get: $\int_0^1 \frac{\sqrt{x} \, dx}{(1+x)(1+3x)(3+x)} = \int_0^1 \frac{u \cdot (2u \, du)}{(1+u^2)(1+3u^2)(3+u^2)} = 2 \int_0^1 \frac{u^2 \, du}{(1+u^2)(1+3u^2)(3+u^2)}$

Step 2: Partial Fraction Decomposition Now we need to decompose the rational function $\frac{u^2}{(1+u^2)(1+3u^2)(3+u^2)}$. Let $y = u^2$. Then the expression becomes $\frac{y}{(1+y)(1+3y)(3+y)}$. We decompose this into partial fractions: $\frac{y}{(1+y)(1+3y)(3+y)} = \frac{A}{1+y} + \frac{B}{1+3y} + \frac{C}{3+y}$ Multiplying both sides by $(1+y)(1+3y)(3+y)$, we get: $y = A(1+3y)(3+y) + B(1+y)(3+y) + C(1+y)(1+3y)$ To find the constants $A$, $B$, and $C$, we can substitute specific values of $y$:

• Let $y = -1$: $-1 = A(1 - 3)(3 - 1) + B(0) + C(0)$ $-1 = A(-2)(2) = -4A \implies A = \frac{1}{4}$
• Let $y = -\frac{1}{3}$: $-\frac{1}{3} = A(0) + B\left(1 - \frac{1}{3}\right)\left(3 - \frac{1}{3}\right) + C(0)$ $-\frac{1}{3} = B\left(\frac{2}{3}\right)\left(\frac{8}{3}\right) = B\left(\frac{16}{9}\right) \implies B = -\frac{1}{3} \cdot \frac{9}{16} = -\frac{3}{16}$
• Let $y = -3$: $-3 = A(0) + B(0) + C(1 - 3)(1 - 9)$ $-3 = C(-2)(-8) = 16C \implies C = -\frac{3}{16}$ So, the partial fraction decomposition is: $\frac{y}{(1+y)(1+3y)(3+y)} = \frac{1/4}{1+y} - \frac{3/16}{1+3y} - \frac{3/16}{3+y}$ Substituting back $y = u^2$: $\frac{u^2}{(1+u^2)(1+3u^2)(3+u^2)} = \frac{1}{4(1+u^2)} - \frac{3}{16(1+3u^2)} - \frac{3}{16(3+u^2)}$

Step 3: Integrating the Partial Fractions Now we integrate the decomposed terms with respect to $u$ from $0$ to $1$. The integral becomes: $2 \int_0^1 \left( \frac{1}{4(1+u^2)} - \frac{3}{16(1+3u^2)} - \frac{3}{16(3+u^2)} \right) du$ We can integrate each term separately:

• Term 1: $\int_0^1 \frac{1}{4(1+u^2)} du = \frac{1}{4} \int_0^1 \frac{1}{1+u^2} du = \frac{1}{4} [\arctan(u)]_0^1 = \frac{1}{4} (\arctan(1) - \arctan(0)) = \frac{1}{4} \left(\frac{\pi}{4} - 0\right) = \frac{\pi}{16}$.
• Term 2: $\int_0^1 \frac{3}{16(1+3u^2)} du = \frac{3}{16} \int_0^1 \frac{1}{1+(\sqrt{3}u)^2} du$. Let $v = \sqrt{3}u$, so $dv = \sqrt{3} du$, which means $du = \frac{1}{\sqrt{3}} dv$. When $u=0$, $v=0$. When $u=1$, $v=\sqrt{3}$. $\frac{3}{16} \int_0^{\sqrt{3}} \frac{1}{1+v^2} \cdot \frac{1}{\sqrt{3}} dv = \frac{3}{16\sqrt{3}} [\arctan(v)]_0^{\sqrt{3}} = \frac{\sqrt{3}}{16} (\arctan(\sqrt{3}) - \arctan(0)) = \frac{\sqrt{3}}{16} \left(\frac{\pi}{3} - 0\right) = \frac{\pi\sqrt{3}}{48}$
• Term 3: $\int_0^1 \frac{3}{16(3+u^2)} du = \frac{3}{16} \int_0^1 \frac{1}{3(1 + u^2/3)} du = \frac{1}{16} \int_0^1 \frac{1}{1+(u/\sqrt{3})^2} du$. Let $w = u/\sqrt{3}$, so $dw = \frac{1}{\sqrt{3}} du$, which means $du = \sqrt{3} dw$. When $u=0$, $w=0$. When $u=1$, $w=1/\sqrt{3}$. $\frac{1}{16} \int_0^{1/\sqrt{3}} \frac{1}{1+w^2} \cdot \sqrt{3} dw = \frac{\sqrt{3}}{16} [\arctan(w)]_0^{1/\sqrt{3}} = \frac{\sqrt{3}}{16} \left(\arctan\left(\frac{1}{\sqrt{3}}\right) - \arctan(0)\right) = \frac{\sqrt{3}}{16} \left(\frac{\pi}{6} - 0\right) = \frac{\pi\sqrt{3}}{96}$

Step 4: Combining the Integrated Terms Now we combine the results of the integration and multiply by the factor of 2 from Step 1. The integral is: $2 \left( \frac{\pi}{16} - \frac{\pi\sqrt{3}}{48} - \frac{\pi\sqrt{3}}{96} \right)$ To combine the terms with $\sqrt{3}$, we find a common denominator: $2 \left( \frac{\pi}{16} - \frac{2\pi\sqrt{3}}{96} - \frac{\pi\sqrt{3}}{96} \right) = 2 \left( \frac{\pi}{16} - \frac{3\pi\sqrt{3}}{96} \right)$ Simplify the fraction $\frac{3}{96}$: $\frac{3}{96} = \frac{1}{32}$. $2 \left( \frac{\pi}{16} - \frac{\pi\sqrt{3}}{32} \right)$ Distribute the 2: $2 \cdot \frac{\pi}{16} - 2 \cdot \frac{\pi\sqrt{3}}{32} = \frac{\pi}{8} - \frac{\pi\sqrt{3}}{16}$ Factor out $\frac{\pi}{8}$: $\frac{\pi}{8} \left( 1 - \frac{\sqrt{3}}{2} \right)$

Common Mistakes & Tips

• Incorrectly changing limits of integration: Always ensure that the limits of integration are correctly transformed according to the substitution made.
• Algebraic errors in partial fraction decomposition: Carefully check the calculations when solving for the constants $A$, $B$, and $C$. A small error here will propagate through the entire solution.
• Errors in standard integral forms: Double-check the formulas for integrals of the form $\int \frac{1}{a^2+u^2} du$ and related forms.
• Simplifying radicals and fractions: Pay close attention to simplifying expressions involving square roots and fractions to avoid errors in the final answer.

Summary

The problem was solved by first applying a substitution $u = \sqrt{x}$ to simplify the integrand and change the limits of integration. This led to a rational function in terms of $u^2$, which was then decomposed into simpler partial fractions. Each of these partial fractions was integrated using standard integral formulas, particularly those involving arctangent. Finally, the results were combined and simplified to obtain the definite value of the integral.

The final answer is $\boxed{{\frac{\pi}{8}\left( {1 - \frac{\sqrt 3 }{2}} \right)}}$, which corresponds to option (A).