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\(4 \int_\limits0^1\left(\frac{1}{\sqrt{3+\mathrm{x}^2}+\sqrt{1+\mathrm{x}^2}}\right) \mathrm{dx}-3 \log _e(\sqrt{3})\) is equal to :

(1) \(2+\sqrt{2}+\log _e(1+\sqrt{2})\)

(2) \(2-\sqrt{2}-\log _e(1+\sqrt{2})\)

(3) \(2+\sqrt{2}-\log _e(1+\sqrt{2})\)

(4) \(2-\sqrt{2}+\log _e(1+\sqrt{2})\)   

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Best answer

Correct option is: (2) \(2-\sqrt{2}-\log _e(1+\sqrt{2})\)    

\(4 \int_\limits0^1 \frac{1}{\sqrt{3+\mathrm{x}^2}+\sqrt{1+\mathrm{x}^2}} \mathrm{dx}-3 ln \sqrt{3}\)   

\(= 4 \int_\limits0^1 \frac{\sqrt{3+x^2}-\sqrt{1+x^2}}{\left(3+\mathrm{x}^2\right)-\left(1-\mathrm{x}^2\right)} \mathrm{dx}-\frac{3}{2} ln 3 \)   

\(= 2\left[\left\{\frac{\mathrm{x}}{2} \sqrt{3+\mathrm{x}^2}+\frac{3}{2} ln \left(\mathrm{x}+\sqrt{3+\mathrm{x}^2}\right)\right\}_0^1\right. \left.-\left\{\frac{\mathrm{x}}{2} \sqrt{1+\mathrm{x}^2}+\frac{1}{2} ln \left(\mathrm{x}+\sqrt{1+\mathrm{x}^2}\right)\right\}_0^1\right]-\frac{3}{2} ln 3 \)   

\(= 2\left[\left\{\frac{1}{2} \sqrt{4}+\frac{3}{2} ln (1+\sqrt{4})\right\}-\left\{0+\frac{3}{2} ln \sqrt{3}\right\}\right. \left.-\left\{\frac{1}{2} \sqrt{2}+\frac{1}{2} ln (1+\sqrt{2})\right\}+\left\{0+\frac{1}{2}(0)\right\}\right]-\frac{3}{2} ln 3\)   

\(=2\left[1+\frac{3}{2} ln 3-\frac{3}{4} ln 3-\frac{1}{\sqrt{2}}-\frac{1}{2} ln (1+\sqrt{2})\right]-\frac{3}{2} ln 3 \)   

\(= 2+3 ln 3-\frac{3}{2} ln 3-\sqrt{2}- ln (1+\sqrt{2})-\frac{3}{2} ln 3 \)   

\(= 2-\sqrt{2}-ln (1+\sqrt{2})\)  

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