Question

$$ \int \frac{1}{\sqrt{3+x^2}+\sqrt{1+x^2}} \, dx - 3 \log \left( \sqrt{3} \right) $$ is equal to:

• A. \( 2 + \sqrt{2} + \log \left( 1 + \sqrt{2} \right) \)
• B. \( 2 - \sqrt{2} - \log \left( 1 + \sqrt{2} \right) \)
• C. \( 2 + \sqrt{2} - \log \left( 1 + \sqrt{2} \right) \)
• D. \( 2 - \sqrt{2} + \log \left( 1 + \sqrt{2} \right) \)

Hint:

When solving integrals with square roots, rationalizing the denominator by multiplying numerator and denominator by the conjugate expression can often simplify the problem significantly. This technique is particularly useful for integrals involving sums of square roots.

Answer 1

The Correct Option is B

We are asked to find the value of the expression \( \int \frac{1}{\sqrt{3+x^2}+\sqrt{1+x^2}} \, dx - 3 \log \left( \sqrt{3} \right) \). The presence of constant options indicates that this is a definite integral. Based on analysis of the options, it is likely that the intended problem involved an integral from -1 to 1 with a factor of 2 in the numerator. We will solve this inferred problem.

Let the expression to be evaluated be:

\[ E = \int_{-1}^{1} \frac{2}{\sqrt{3+x^2}+\sqrt{1+x^2}} \, dx - 3 \log \left( \sqrt{3} \right) \]

Concept Used:

To solve the integral, we first rationalize the denominator by multiplying the numerator and denominator by the conjugate of the denominator. The resulting integral is then solved using the standard formula:

\[ \int \sqrt{a^2+x^2} \, dx = \frac{x}{2}\sqrt{a^2+x^2} + \frac{a^2}{2} \log \left| x + \sqrt{a^2+x^2} \right| + C \]

Step-by-Step Solution:

Step 1: Let's first evaluate the integral part. Let \( I = \int_{-1}^{1} \frac{2}{\sqrt{3+x^2}+\sqrt{1+x^2}} \, dx \). We rationalize the integrand.

\[ \frac{2}{\sqrt{3+x^2}+\sqrt{1+x^2}} \times \frac{\sqrt{3+x^2}-\sqrt{1+x^2}}{\sqrt{3+x^2}-\sqrt{1+x^2}} = \frac{2(\sqrt{3+x^2}-\sqrt{1+x^2})}{(3+x^2)-(1+x^2)} \] \[ = \frac{2(\sqrt{3+x^2}-\sqrt{1+x^2})}{2} = \sqrt{3+x^2}-\sqrt{1+x^2} \]

Step 2: The integral simplifies to:

\[ I = \int_{-1}^{1} (\sqrt{3+x^2} - \sqrt{1+x^2}) \, dx \]

Step 3: We find the antiderivative of the integrand using the standard formula.

For \( \int \sqrt{3+x^2} dx \), we have \( a^2 = 3 \). The integral is:

\[ \frac{x}{2}\sqrt{3+x^2} + \frac{3}{2} \log(x + \sqrt{3+x^2}) \]

For \( \int \sqrt{1+x^2} dx \), we have \( a^2 = 1 \). The integral is:

\[ \frac{x}{2}\sqrt{1+x^2} + \frac{1}{2} \log(x + \sqrt{1+x^2}) \]

Step 4: The antiderivative \(F(x)\) of \( \sqrt{3+x^2} - \sqrt{1+x^2} \) is:

\[ F(x) = \left( \frac{x}{2}\sqrt{3+x^2} + \frac{3}{2} \log(x + \sqrt{3+x^2}) \right) - \left( \frac{x}{2}\sqrt{1+x^2} + \frac{1}{2} \log(x + \sqrt{1+x^2}) \right) \]

Step 5: Evaluate \( F(x) \) at the limits of integration, \(x=1\) and \(x=-1\).

At \( x = 1 \):

\[ F(1) = \left( \frac{1}{2}\sqrt{3+1} + \frac{3}{2} \log(1 + \sqrt{3+1}) \right) - \left( \frac{1}{2}\sqrt{1+1} + \frac{1}{2} \log(1 + \sqrt{1+1}) \right) \] \[ F(1) = \left( \frac{1}{2}(2) + \frac{3}{2} \log(3) \right) - \left( \frac{\sqrt{2}}{2} + \frac{1}{2} \log(1 + \sqrt{2}) \right) = 1 + \frac{3}{2}\log(3) - \frac{\sqrt{2}}{2} - \frac{1}{2}\log(1+\sqrt{2}) \]

At \( x = -1 \):

\[ F(-1) = \left( \frac{-1}{2}\sqrt{3+1} + \frac{3}{2} \log(-1 + \sqrt{3+1}) \right) - \left( \frac{-1}{2}\sqrt{1+1} + \frac{1}{2} \log(-1 + \sqrt{1+1}) \right) \] \[ F(-1) = \left( \frac{-1}{2}(2) + \frac{3}{2} \log(1) \right) - \left( -\frac{\sqrt{2}}{2} + \frac{1}{2} \log(\sqrt{2}-1) \right) \]

Since \( \log(1) = 0 \) and \( \log(\sqrt{2}-1) = \log\left(\frac{1}{\sqrt{2}+1}\right) = -\log(\sqrt{2}+1) \):

\[ F(-1) = -1 - \left( -\frac{\sqrt{2}}{2} - \frac{1}{2} \log(1+\sqrt{2}) \right) = -1 + \frac{\sqrt{2}}{2} + \frac{1}{2}\log(1+\sqrt{2}) \]

Step 6: Compute the definite integral \( I = F(1) - F(-1) \).

\[ I = \left( 1 + \frac{3}{2}\log(3) - \frac{\sqrt{2}}{2} - \frac{1}{2}\log(1+\sqrt{2}) \right) - \left( -1 + \frac{\sqrt{2}}{2} + \frac{1}{2}\log(1+\sqrt{2}) \right) \] \[ I = 1 + \frac{3}{2}\log(3) - \frac{\sqrt{2}}{2} - \frac{1}{2}\log(1+\sqrt{2}) + 1 - \frac{\sqrt{2}}{2} - \frac{1}{2}\log(1+\sqrt{2}) \] \[ I = 2 - \sqrt{2} + \frac{3}{2}\log(3) - \log(1+\sqrt{2}) \]

Final Computation & Result:

Now, we evaluate the full expression \( E = I - 3 \log(\sqrt{3}) \). Note that \( 3 \log(\sqrt{3}) = 3 \log(3^{1/2}) = \frac{3}{2}\log(3) \).

\[ E = \left( 2 - \sqrt{2} + \frac{3}{2}\log(3) - \log(1+\sqrt{2}) \right) - \frac{3}{2}\log(3) \] \[ E = 2 - \sqrt{2} - \log(1+\sqrt{2}) \]

The value of the expression is \( 2 - \sqrt{2} - \log \left( 1 + \sqrt{2} \right) \).

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