Definite Integrals Part 3

The value is $4\int\limits_{0}^{1}\frac{dx}{\sqrt{3+x^2}+\sqrt{1+x^2}} - 3\log_e\sqrt{3}$.Rationalise the denominator by multiplying numerator and denominator by $\sqrt{3+x^2}-\sqrt{1+x^2}$:$\frac{1}{\sqrt{3+x^2}+\sqrt{1+x^2}} = \frac{\sqrt{3+x^2}-\sqrt{1+x^2}}{(3+x^2)-(1+x^2)} = \frac{\sqrt{3+x^2}-\sqrt{1+x^2}}{2}$.Hence the expression equals $2\int\limits_{0}^{1}\left(\sqrt{3+x^2}-\sqrt{1+x^2}\right)dx - 3\log_e\sqrt{3}$.Use the standard result $\int \sqrt{x^2+a^2}\,dx = \frac{x}{2}\sqrt{x^2+a^2} + \frac{a^2}{2}\ln\left(x+\sqrt{x^2+a^2}\right)$.For $a^2=3$:$\int\limits_{0}^{1}\sqrt{3+x^2}\,dx = \left[\frac{x}{2}\sqrt{x^2+3}+\frac{3}{2}\ln\left(x+\sqrt{x^2+3}\right)\right]_{0}^{1} = 1+\frac{3}{2}\ln 3-\frac{3}{2}\ln\sqrt{3} = 1+\frac{3}{4}\ln 3$.For $a^2=1$:$\int\limits_{0}^{1}\sqrt{1+x^2}\,dx = \left[\frac{x}{2}\sqrt{x^2+1}+\frac{1}{2}\ln\left(x+\sqrt{x^2+1}\right)\right]_{0}^{1} = \frac{\sqrt{2}}{2}+\frac{1}{2}\ln(1+\sqrt{2})$.Therefore the given expression equals$2\left[\left(1+\frac{3}{4}\ln 3\right)-\left(\frac{\sqrt{2}}{2}+\frac{1}{2}\ln(1+\sqrt{2})\right)\right]-3\log_e\sqrt{3}$$= 2+\frac{3}{2}\ln 3-\sqrt{2}-\ln(1+\sqrt{2})-\frac{3}{2}\ln 3$.The $\ln 3$ terms cancel since $3\log_e\sqrt{3}=\frac{3}{2}\ln 3$:$= 2-\sqrt{2}-\ln(1+\sqrt{2})$.Hence the required value is $2-\sqrt{2}-\log_e(1+\sqrt{2})$, which matches option B.The option marked as correct (option C) is wrong because it has the sign of $\log_e(1+\sqrt{2})$ reversed.