After rationalizing, the denominator simplifies to (3 + x²) - (1 + x²) = 3 + x² - 1 - x² = 2. We can then factor out the constant (1)/(2) from the integral.

We use the standard integral formula for ∫ √(a² + x²) dx. For the first term, a² = 3, so a = √3. For the second term, a² = 1, so a = 1.

Simplify the terms. We have log_e(3) = 2log_e(√3).

$4\int_0^1 \left(\frac{1}{\sqrt{3 + x^2} + \sqrt{1 + x^2}}\right) dx - 3 \log_e(\sqrt{3})$ is equal to:

Answer: $2 - \sqrt{2} - \log_e(1 + \sqrt{2})$

Step-by-step solution

Step 1: Rationalize the integrand

To simplify the integrand, we rationalize the denominator by multiplying the numerator and denominator by the conjugate of the denominator, which is $(\sqrt{3 + x^2} - \sqrt{1 + x^2})$ . This uses the difference of squares formula, $(a+b)(a-b) = a^2 - b^2$ .

Step 2: Simplify the integrand

After rationalizing, the denominator simplifies to $(3 + x^2) - (1 + x^2) = 3 + x^2 - 1 - x^2 = 2$ . We can then factor out the constant $\frac{1}{2}$ from the integral.

Step 3: Apply the standard integral formula

We use the standard integral formula for $\int \sqrt{a^2 + x^2} dx$ . For the first term, $a^2 = 3$ , so $a = \sqrt{3}$ . For the second term, $a^2 = 1$ , so $a = 1$ .

Step 4: Evaluate the definite integral

Now we apply the integral formula to both terms and evaluate the definite integral from $0$ to $1$ . We substitute the upper limit $x=1$ and subtract the value obtained by substituting the lower limit $x=0$ .

Step 5: Calculate the value at limits

Substitute $x=1$ and $x=0$ into the integrated expression. Note that $\sqrt{4} = 2$ and $\log_e|1| = 0$ .

Step 6: Simplify the integral result

Simplify the terms. We have $\log_e(3) = 2\log_e(\sqrt{3})$ .

Step 7: Calculate the final expression

Substitute the value of $I$ back into the original expression and simplify. Combine the logarithmic terms.

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4∫01(13+x2+1+x2)dx−3log⁡e(3)4\int_0^1 \left(\frac{1}{\sqrt{3 + x^2} + \sqrt{1 + x^2}}\right) dx - 3 \log_e(\sqrt{3})4∫01​(3+x2​+1+x2​1​)dx−3loge​(3​) is equal to: