1. Key Concepts and Formulas

• Rationalization of Denominators: To simplify expressions involving square roots in the denominator, multiply the numerator and denominator by the conjugate of the expression. This uses the identity $(A+B)(A-B) = A^2 - B^2$.
• Power Rule of Integration: The integral of $x^n$ is $\frac{x^{n+1}}{n+1} + C$ for $n \neq -1$. This rule is extended to linear functions of the form $(ax+b)^n$.
• Definite Integral Evaluation: The definite integral $\int_p^q f(x) \,dx$ is evaluated by finding an antiderivative $F(x)$ of $f(x)$ and computing $F(q) - F(p)$.
• Comparison of Coefficients: When an expression is given in the form $a+b\sqrt{2}+c\sqrt{3}$, where $a, b, c$ are rational numbers, we can equate coefficients of the rational part, $\sqrt{2}$, and $\sqrt{3}$ from the calculated integral to find the values of $a, b, c$.

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2. Step-by-Step Solution

Let the given integral be $I$. $I = \int\limits_0^1 \frac{1}{\sqrt{3+x}+\sqrt{1+x}} \mathrm{~d} x$

Step 1: Rationalize the Integrand To simplify the integrand, we multiply the numerator and the denominator by the conjugate of the denominator, which is $\sqrt{3+x}-\sqrt{1+x}$. This is done to eliminate the square roots in the denominator. $I = \int\limits_0^1 \frac{1}{\sqrt{3+x}+\sqrt{1+x}} \cdot \frac{\sqrt{3+x}-\sqrt{1+x}}{\sqrt{3+x}-\sqrt{1+x}} \mathrm{~d} x$ Using the difference of squares formula $(A+B)(A-B) = A^2 - B^2$ for the denominator: $I = \int\limits_0^1 \frac{\sqrt{3+x}-\sqrt{1+x}}{(\sqrt{3+x})^2 - (\sqrt{1+x})^2} \mathrm{~d} x$ $I = \int\limits_0^1 \frac{\sqrt{3+x}-\sqrt{1+x}}{(3+x) - (1+x)} \mathrm{~d} x$ Simplify the denominator: $(3+x) - (1+x) = 3+x-1-x = 2$. $I = \int\limits_0^1 \frac{\sqrt{3+x}-\sqrt{1+x}}{2} \mathrm{~d} x$

Step 2: Separate and Integrate the Terms We can factor out the constant $\frac{1}{2}$ and use the linearity of integration to split the integral into two parts. The integrand is now in a form that can be integrated using the power rule. $I = \frac{1}{2} \int\limits_0^1 \left( (3+x)^{1/2} - (1+x)^{1/2} \right) \mathrm{~d} x$ $I = \frac{1}{2} \left[ \int\limits_0^1 (3+x)^{1/2} \mathrm{~d} x - \int\limits_0^1 (1+x)^{1/2} \mathrm{~d} x \right]$ We apply the power rule for integration, $\int (ax+b)^n \,dx = \frac{(ax+b)^{n+1}}{a(n+1)}$. Here, $a=1$ and $n=1/2$. For the first integral: $\int (3+x)^{1/2} \mathrm{~d} x = \frac{(3+x)^{1/2+1}}{1/2+1} = \frac{(3+x)^{3/2}}{3/2} = \frac{2}{3}(3+x)^{3/2}$. For the second integral: $\int (1+x)^{1/2} \mathrm{~d} x = \frac{(1+x)^{1/2+1}}{1/2+1} = \frac{(1+x)^{3/2}}{3/2} = \frac{2}{3}(1+x)^{3/2}$. Substituting these back into the expression for $I$: $I = \frac{1}{2} \left[ \frac{2}{3}(3+x)^{3/2} - \frac{2}{3}(1+x)^{3/2} \right]_0^1$ Factor out $\frac{2}{3}$: $I = \frac{1}{2} \cdot \frac{2}{3} \left[ (3+x)^{3/2} - (1+x)^{3/2} \right]_0^1$ $I = \frac{1}{3} \left[ (3+x)^{3/2} - (1+x)^{3/2} \right]_0^1$

Step 3: Evaluate the Definite Integral Now, we substitute the upper limit ($x=1$) and the lower limit ($x=0$) into the integrated expression and subtract. $I = \frac{1}{3} \left[ \left( (3+1)^{3/2} - (1+1)^{3/2} \right) - \left( (3+0)^{3/2} - (1+0)^{3/2} \right) \right]$ $I = \frac{1}{3} \left[ \left( 4^{3/2} - 2^{3/2} \right) - \left( 3^{3/2} - 1^{3/2} \right) \right]$ Calculate the powers:

• $4^{3/2} = (\sqrt{4})^3 = 2^3 = 8$
• $2^{3/2} = (\sqrt{2})^3 = 2\sqrt{2}$
• $3^{3/2} = (\sqrt{3})^3 = 3\sqrt{3}$
• $1^{3/2} = 1$ Substitute these values back: $I = \frac{1}{3} \left[ (8 - 2\sqrt{2}) - (3\sqrt{3} - 1) \right]$

Step 4: Simplify the Resulting Expression Distribute the negative sign and combine like terms. $I = \frac{1}{3} \left[ 8 - 2\sqrt{2} - 3\sqrt{3} + 1 \right]$ $I = \frac{1}{3} \left[ 9 - 2\sqrt{2} - 3\sqrt{3} \right]$ Distribute the $\frac{1}{3}$ to each term: $I = \frac{9}{3} - \frac{2}{3}\sqrt{2} - \frac{3}{3}\sqrt{3}$ $I = 3 - \frac{2}{3}\sqrt{2} - \sqrt{3}$

Step 5: Determine the Values of a, b, and c The problem states that the integral is equal to $a+b\sqrt{2}+c\sqrt{3}$, where $a, b, c$ are rational numbers. By comparing our result with this form:

• The rational part is $3$, so $a=3$.
• The coefficient of $\sqrt{2}$ is $-\frac{2}{3}$, so $b=-\frac{2}{3}$.
• The coefficient of $\sqrt{3}$ is $-1$, so $c=-1$. All these values are rational numbers, as required.

Step 6: Calculate the Final Expression We need to find the value of $2a+3b-4c$. Substitute the determined values of $a, b, c$: $2a+3b-4c = 2(3) + 3\left(-\frac{2}{3}\right) - 4(-1)$ $2a+3b-4c = 6 - 2 + 4$ $2a+3b-4c = 4 + 4$ $2a+3b-4c = 8$

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3. Common Mistakes & Tips

• Sign Errors during Rationalization: Be meticulous when expanding the denominator after multiplying by the conjugate. Ensure all terms are correctly squared and subtracted.
• Incorrect Power Rule Application: Double-check the exponent calculation ($n+1$) and the division by $a(n+1)$ when integrating expressions like $(ax+b)^n$.
• Arithmetic Mistakes in Evaluation: When substituting the limits of integration, carefully calculate powers of numbers, especially fractional exponents, and simplify the resulting arithmetic expression.

4. Summary

The given definite integral was solved by first rationalizing the integrand to simplify it. The resulting expression was then integrated using the power rule for integration. After evaluating the definite integral at the limits and simplifying, the result was compared with the given form $a+b\sqrt{2}+c\sqrt{3}$ to find the rational coefficients $a, b,$ and $c$. Finally, the required expression $2a+3b-4c$ was calculated using these coefficients.

5. Final Answer

The final answer is \boxed{8} which corresponds to option (D).