Key Concepts and Formulas

• Rationalization of Denominators: To simplify expressions involving square roots in the denominator, we multiply the numerator and denominator by the conjugate of the denominator. For an expression of the form $\frac{1}{\sqrt{a}+\sqrt{b}}$, the conjugate is $\sqrt{a}-\sqrt{b}$. This uses the identity $(x+y)(x-y) = x^2-y^2$.
• Standard Integral Formula: The integral of $\frac{1}{\sqrt{a^2+x^2}}$ is a standard form: $\int \frac{1}{\sqrt{a^2+x^2}} \, dx = \log_e |x + \sqrt{a^2+x^2}| + C$.
• Definite Integration Properties: The definite integral $\int_a^b f(x) \, dx$ is evaluated using the Fundamental Theorem of Calculus: $\int_a^b f(x) \, dx = F(b) - F(a)$, where $F(x)$ is an antiderivative of $f(x)$.

Step-by-Step Solution

We are asked to evaluate $4 \int_0^1\left(\frac{1}{\sqrt{3+x^2}+\sqrt{1+x^2}}\right) d x-3 \log _e(\sqrt{3})$.

Step 1: Simplify the integrand by rationalizing the denominator. The integrand is $\frac{1}{\sqrt{3+x^2}+\sqrt{1+x^2}}$. To simplify this, we multiply the numerator and denominator by the conjugate of the denominator, which is $\sqrt{3+x^2}-\sqrt{1+x^2}$. $\frac{1}{\sqrt{3+x^2}+\sqrt{1+x^2}} \times \frac{\sqrt{3+x^2}-\sqrt{1+x^2}}{\sqrt{3+x^2}-\sqrt{1+x^2}} = \frac{\sqrt{3+x^2}-\sqrt{1+x^2}}{(\sqrt{3+x^2})^2 - (\sqrt{1+x^2})^2}$ $= \frac{\sqrt{3+x^2}-\sqrt{1+x^2}}{(3+x^2) - (1+x^2)} = \frac{\sqrt{3+x^2}-\sqrt{1+x^2}}{3+x^2-1-x^2} = \frac{\sqrt{3+x^2}-\sqrt{1+x^2}}{2}$ So, the integrand simplifies to $\frac{1}{2}(\sqrt{3+x^2}-\sqrt{1+x^2})$.

Step 2: Evaluate the definite integral of the simplified integrand. Now we need to compute $\int_0^1 \frac{1}{2}(\sqrt{3+x^2}-\sqrt{1+x^2}) \, dx$. This can be split into two separate integrals: $\frac{1}{2} \int_0^1 \sqrt{3+x^2} \, dx - \frac{1}{2} \int_0^1 \sqrt{1+x^2} \, dx$ We use the standard integral formula $\int \sqrt{a^2+x^2} \, dx = \frac{x}{2}\sqrt{a^2+x^2} + \frac{a^2}{2}\log_e|x+\sqrt{a^2+x^2}| + C$.

For the first integral, $\int \sqrt{3+x^2} \, dx$, we have $a^2=3$, so $a=\sqrt{3}$. $\int_0^1 \sqrt{3+x^2} \, dx = \left[ \frac{x}{2}\sqrt{3+x^2} + \frac{3}{2}\log_e|x+\sqrt{3+x^2}| \right]_0^1$ $= \left( \frac{1}{2}\sqrt{3+1^2} + \frac{3}{2}\log_e|1+\sqrt{3+1^2}| \right) - \left( \frac{0}{2}\sqrt{3+0^2} + \frac{3}{2}\log_e|0+\sqrt{3+0^2}| \right)$ $= \left( \frac{1}{2}\sqrt{4} + \frac{3}{2}\log_e|1+\sqrt{4}| \right) - \left( 0 + \frac{3}{2}\log_e|\sqrt{3}| \right)$ $= \left( \frac{1}{2}(2) + \frac{3}{2}\log_e(1+2) \right) - \frac{3}{2}\log_e(\sqrt{3}) = \left( 1 + \frac{3}{2}\log_e(3) \right) - \frac{3}{2}\log_e(\sqrt{3})$ $= 1 + \frac{3}{2}\log_e(3) - \frac{3}{2}\log_e(3^{1/2}) = 1 + \frac{3}{2}\log_e(3) - \frac{3}{2} \times \frac{1}{2}\log_e(3) = 1 + \frac{3}{2}\log_e(3) - \frac{3}{4}\log_e(3)$ $= 1 + \left(\frac{3}{2} - \frac{3}{4}\right)\log_e(3) = 1 + \frac{3}{4}\log_e(3)$

For the second integral, $\int \sqrt{1+x^2} \, dx$, we have $a^2=1$, so $a=1$. $\int_0^1 \sqrt{1+x^2} \, dx = \left[ \frac{x}{2}\sqrt{1+x^2} + \frac{1}{2}\log_e|x+\sqrt{1+x^2}| \right]_0^1$ $= \left( \frac{1}{2}\sqrt{1+1^2} + \frac{1}{2}\log_e|1+\sqrt{1+1^2}| \right) - \left( \frac{0}{2}\sqrt{1+0^2} + \frac{1}{2}\log_e|0+\sqrt{1+0^2}| \right)$ $= \left( \frac{1}{2}\sqrt{2} + \frac{1}{2}\log_e|1+\sqrt{2}| \right) - \left( 0 + \frac{1}{2}\log_e(1) \right)$ $= \frac{\sqrt{2}}{2} + \frac{1}{2}\log_e(1+\sqrt{2}) - 0 = \frac{\sqrt{2}}{2} + \frac{1}{2}\log_e(1+\sqrt{2})$

Now, substitute these back into the expression from Step 2: $\frac{1}{2} \left( 1 + \frac{3}{4}\log_e(3) \right) - \frac{1}{2} \left( \frac{\sqrt{2}}{2} + \frac{1}{2}\log_e(1+\sqrt{2}) \right)$ $= \frac{1}{2} + \frac{3}{8}\log_e(3) - \frac{\sqrt{2}}{4} - \frac{1}{4}\log_e(1+\sqrt{2})$

The original question asks for $4 \int_0^1\left(\frac{1}{\sqrt{3+x^2}+\sqrt{1+x^2}}\right) d x-3 \log _e(\sqrt{3})$. We have $4 \times \frac{1}{2} \left( \int_0^1 \sqrt{3+x^2} \, dx - \int_0^1 \sqrt{1+x^2} \, dx \right)$. So, $2 \left( \left( 1 + \frac{3}{4}\log_e(3) \right) - \left( \frac{\sqrt{2}}{2} + \frac{1}{2}\log_e(1+\sqrt{2}) \right) \right)$. $= 2 + \frac{3}{2}\log_e(3) - \sqrt{2} - \log_e(1+\sqrt{2})$

Now, we need to subtract $3 \log_e(\sqrt{3})$. $3 \log_e(\sqrt{3}) = 3 \log_e(3^{1/2}) = 3 \times \frac{1}{2} \log_e(3) = \frac{3}{2} \log_e(3)$

So the complete expression is: $\left( 2 + \frac{3}{2}\log_e(3) - \sqrt{2} - \log_e(1+\sqrt{2}) \right) - \frac{3}{2}\log_e(3)$ $= 2 - \sqrt{2} - \log_e(1+\sqrt{2})$

Common Mistakes & Tips

• Algebraic Errors: Be careful with the signs and exponents when rationalizing the denominator and applying the formula for $\int \sqrt{a^2+x^2} \, dx$.
• Logarithm Properties: Remember that $\log_e(\sqrt{a}) = \frac{1}{2}\log_e(a)$. Also, $\log_e(1) = 0$.
• Trigonometric Substitution: While not used here, for integrals involving $\sqrt{a^2+x^2}$, trigonometric substitution ($x=a\tan\theta$) is another valid method, but often more cumbersome than using the standard formula.

Summary

The problem involves evaluating a definite integral of a rationalized expression involving square roots and then subtracting a logarithmic term. We first rationalized the denominator of the integrand to simplify it. Then, we applied the standard formula for integrating $\sqrt{a^2+x^2}$ to evaluate the definite integral over the given limits. Finally, we combined the results and subtracted the given logarithmic term, leading to the simplified expression.

The final answer is \boxed{2-\sqrt{2}-\log _{\mathrm{e}}(1+\sqrt{2})}. This corresponds to option (A).