$4\int_{0}^{1} (\frac{1}{\sqrt{3+x^2} + \sqrt{1+x^2}}) dx - 3\log_e (\sqrt{3})$ is equal to :
The problem asks for the evaluation of the expression: $ 4\int_{0}^{1} \left( \frac{1}{\sqrt{3+x^2} + \sqrt{1+x^2}} \right) dx - 3\log_e (\sqrt{3}) $
Simplify the Integrand
First, simplify the fraction inside the integral by multiplying the numerator and the 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}} = \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} $
Evaluate the Definite Integral
The integral part of the expression becomes: $ 4\int_{0}^{1} \frac{\sqrt{3+x^2} - \sqrt{1+x^2}}{2} dx = 2 \int_{0}^{1} (\sqrt{3+x^2} - \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}) $. Let's evaluate $ \int_{0}^{1} \sqrt{3+x^2} dx $ (where $a^2=3$): $ \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( 0 + \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) - \frac{3}{2}\log_e(\sqrt{3}) $ $ = \left( \frac{1}{2}(2) + \frac{3}{2}\log_e(1+2) \right) - \frac{3}{2}\log_e(\sqrt{3}) $ $ = 1 + \frac{3}{2}\log_e(3) - \frac{3}{2}\log_e(\sqrt{3}) = 1 + \frac{3}{2} \log_e\left(\frac{3}{\sqrt{3}}\right) = 1 + \frac{3}{2}\log_e(\sqrt{3}) $ Now, evaluate $ \int_{0}^{1} \sqrt{1+x^2} dx $ (where $a^2=1$): $ \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( 0 + \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) - \frac{1}{2}\log_e(1) $ $ = \frac{\sqrt{2}}{2} + \frac{1}{2}\log_e(1+\sqrt{2}) \quad (\text{since } \log_e(1)=0) $ The integral part is: $ 2 \left[ \left( 1 + \frac{3}{2}\log_e(\sqrt{3}) \right) - \left( \frac{\sqrt{2}}{2} + \frac{1}{2}\log_e(1+\sqrt{2}) \right) \right] $ $ = 2 \left[ 1 - \frac{\sqrt{2}}{2} + \frac{3}{2}\log_e(\sqrt{3}) - \frac{1}{2}\log_e(1+\sqrt{2}) \right] $ $ = 2 - \sqrt{2} + 3\log_e(\sqrt{3}) - \log_e(1+\sqrt{2}) $
Final Simplification
Now, substitute this back into the original expression: $ \left( 2 - \sqrt{2} + 3\log_e(\sqrt{3}) - \log_e(1+\sqrt{2}) \right) - 3\log_e(\sqrt{3}) $ The terms $ 3\log_e(\sqrt{3}) $ cancel out. $ = 2 - \sqrt{2} - \log_e(1+\sqrt{2}) $
Result
The value of the expression is $ 2-\sqrt{2}-\log_e (1+\sqrt{2}) $. This corresponds to Option B.
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