JEE Main 2025 Apr 2 Shift 2, Mathematics Q13: Integrals of Some Particular Functions

Rationalise the integrand by multiplying numerator and denominator by $\sqrt{3+x^2}-\sqrt{1+x^2}$:

$\frac{1}{\sqrt{3+x^2}+\sqrt{1+x^2}}=\frac{\sqrt{3+x^2}-\sqrt{1+x^2}}{(3+x^2)-(1+x^2)}=\frac{\sqrt{3+x^2}-\sqrt{1+x^2}}{2}$.

So the expression becomes $2{\int }_0^1\left(\sqrt{3+x^2}-\sqrt{1+x^2}\right)dx-\frac{3}{2}\ln 3$.

Standard antiderivative: $\int \sqrt{a^2+x^2}dx=\frac{x}{2}\sqrt{a^2+x^2}+\frac{a^2}{2}\ln \left(x+\sqrt{a^2+x^2}\right)$.

For $a^2=3$ on $[0,1]$: $\left[\frac{1}{2}\cdot 2+\frac{3}{2}\ln (1+2)-0-\frac{3}{2}\ln \sqrt{3}\right]=1+\frac{3}{2}\ln 3-\frac{3}{4}\ln 3=1+\frac{3}{4}\ln 3$.

For $a^2=1$ on $[0,1]$: $\left[\frac{1}{2}\sqrt{2}+\frac{1}{2}\ln (1+\sqrt{2})-0\right]=\frac{\sqrt{2}}{2}+\frac{1}{2}\ln (1+\sqrt{2})$.

Doubling the difference: $2\left[1+\frac{3}{4}\ln 3-\frac{\sqrt{2}}{2}-\frac{1}{2}\ln (1+\sqrt{2})\right]=2+\frac{3}{2}\ln 3-\sqrt{2}-\ln (1+\sqrt{2})$.

Subtracting $\frac{3}{2}\ln 3$ gives $2-\sqrt{2}-\ln (1+\sqrt{2})$.