\(4{\int }_{0}^{1}\left(\frac{1}{\sqrt{3+{\mathrm{x}}^{2}}+\sqrt{1+{\mathrm{x}}^{2}}}\right)\mathrm{dx}-3{\log }_{\mathr…

Let \(I=4\int_0^1\left(\frac{1}{\sqrt{3+x^2}+\sqrt{1+x^2}}\right)dx-3\log_e(\sqrt{3})\).

\(I=2\int_0^1\left(\sqrt{x^2+3}-\sqrt{x^2+1}\right)dx-3\log_e(\sqrt{3})\).

Now \( \int \sqrt{x^2+a^2}dx=\frac{x}{2}\sqrt{x^2+a^2}+\frac{a^2}{2}\log_e\left(x+\sqrt{x^2+a^2}\right) \).

Therefore \( \int_0^1 \sqrt{x^2+3}dx=\left[\frac{x}{2}\sqrt{x^2+3}+\frac{3}{2}\log_e\left(x+\sqrt{x^2+3}\right)\right]_0^1 \).

So \( \int_0^1 \sqrt{x^2+3}dx=1+\frac{3}{2}\log_e\left(\frac{3}{\sqrt{3}}\right)=1+\frac{3}{2}\log_e(\sqrt{3}) \).

Also \( \int_0^1 \sqrt{x^2+1}dx=\left[\frac{x}{2}\sqrt{x^2+1}+\frac{1}{2}\log_e\left(x+\sqrt{x^2+1}\right)\right]_0^1 \).

So \( \int_0^1 \sqrt{x^2+1}dx=\frac{\sqrt{2}}{2}+\frac{1}{2}\log_e(1+\sqrt{2}) \).

Hence \(I=2\left(1+\frac{3}{2}\log_e(\sqrt{3})-\frac{\sqrt{2}}{2}-\frac{1}{2}\log_e(1+\sqrt{2})\right)-3\log_e(\sqrt{3})\).

So \(I=2-\sqrt{2}+3\log_e(\sqrt{3})-\log_e(1+\sqrt{2})-3\log_e(\sqrt{3})\).

Therefore \(I=2-\sqrt{2}-\log_e(1+\sqrt{2})\).