4 int_0^1 left(frac{1}{sqrt{3+x^2}+sqrt{1+x^2 | vedclass

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(B) Let $I = 4 \int_0^1 \frac{1}{\sqrt{3+x^2}+\sqrt{1+x^2}} dx - 3 \ln \sqrt{3}$.Rationalizing the integrand:$I = 4 \int_0^1 \frac{\sqrt{3+x^2}-\sqrt{

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

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(B) Let $I = 4 \int_0^1 \frac{1}{\sqrt{3+x^2}+\sqrt{1+x^2}} dx - 3 \ln \sqrt{3}$.Rationalizing the integrand:$I = 4 \int_0^1 \frac{\sqrt{3+x^2}-\sqrt{1+x^2}}{(3+x^2)-(1+x^2)} dx - \frac{3}{2} \ln 3$$I = 4 \int_0^1 \frac{\sqrt{3+x^2}-\sqrt{1+x^2}}{2} dx - \frac{3}{2} \ln 3$$I = 2 \int_0^1 \sqrt{3+x^2} dx - 2 \int_0^1 \sqrt{1+x^2} dx - \frac{3}{2} \ln 3$Using the formula $\int \sqrt{a^2+x^2} dx = \frac{x}{2}\sqrt{a^2+x^2} + \frac{a^2}{2}\ln(x+\sqrt{a^2+x^2})$:$I = 2 \left[ \frac{x}{2}\sqrt{3+x^2} + \frac{3}{2}\ln(x+\sqrt{3+x^2}) \right]_0^1 - 2 \left[ \frac{x}{2}\sqrt{1+x^2} + \frac{1}{2}\ln(x+\sqrt{1+x^2}) \right]_0^1 - \frac{3}{2} \ln 3$$I = \left[ x\sqrt{3+x^2} + 3\ln(x+\sqrt{3+x^2}) \right]_0^1 - \left[ x\sqrt{1+x^2} + \ln(x+\sqrt{1+x^2}) \right]_0^1 - \frac{3}{2} \ln 3$$I = (1\sqrt{4} + 3\ln(1+2) - (0 + 3\ln\sqrt{3})) - (1\sqrt{2} + \ln(1+\sqrt{2}) - (0 + \ln 1)) - \frac{3}{2} \ln 3$$I = (2 + 3\ln 3 - \frac{3}{2}\ln 3) - (\sqrt{2} + \ln(1+\sqrt{2})) - \frac{3}{2} \ln 3$$I = 2 + \frac{3}{2}\ln 3 - \sqrt{2} - \ln(1+\sqrt{2}) - \frac{3}{2} \ln 3$$I = 2 - \sqrt{2} - \ln(1+\sqrt{2})$.

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

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