Integral Calculus – Definite Integration – JEE Main 02 April 2025 Shift 2

Solution:

Rationalize the denominator of the integrand:
$$4\int_{0}^{1}\frac{\sqrt{3+x^{2}}-\sqrt{1+x^{2}}}{(\sqrt{3+x^{2}}+\sqrt{1+x^{2}})(\sqrt{3+x^{2}}-\sqrt{1+x^{2}})}dx – 3\log_{e}(\sqrt{3})$$

$$4\int_{0}^{1}\frac{\sqrt{3+x^{2}}-\sqrt{1+x^{2}}}{(3+x^{2})-(1+x^{2})}dx – 3\log_{e}(3^{1/2})$$

$$4\int_{0}^{1}\frac{\sqrt{3+x^{2}}-\sqrt{1+x^{2}}}{2}dx – \frac{3}{2}\log_{e}(3)$$

$$2\int_{0}^{1}\sqrt{3+x^{2}}dx – 2\int_{0}^{1}\sqrt{1+x^{2}}dx – \frac{3}{2}\log_{e}(3)$$

Using the standard integration formula $\int \sqrt{x^2+a^2} dx = \frac{x}{2}\sqrt{x^2+a^2} + \frac{a^2}{2}\log_{e}|x+\sqrt{x^2+a^2}|$:

$$2 \left[ \frac{x}{2}\sqrt{x^{2}+3} + \frac{3}{2}\log_{e}|x+\sqrt{x^{2}+3}| \right]_{0}^{1} – 2 \left[ \frac{x}{2}\sqrt{x^{2}+1} + \frac{1}{2}\log_{e}|x+\sqrt{x^{2}+1}| \right]_{0}^{1} – \frac{3}{2}\log_{e}(3)$$

Substitute the upper limit ($1$) and lower limit ($0$):
$$2 \left[ \left(\frac{1}{2}\sqrt{4} + \frac{3}{2}\log_{e}(1+\sqrt{4})\right) – \left(0 + \frac{3}{2}\log_{e}(\sqrt{3})\right) \right] – 2 \left[ \left(\frac{1}{2}\sqrt{2} + \frac{1}{2}\log_{e}(1+\sqrt{2})\right) – (0) \right] – \frac{3}{2}\log_{e}(3)$$

$$2 \left[ \left(1 + \frac{3}{2}\log_{e}(3)\right) – \left(\frac{3}{4}\log_{e}(3)\right) \right] – \left[ \sqrt{2} + \log_{e}(1+\sqrt{2}) \right] – \frac{3}{2}\log_{e}(3)$$

$$2 \left[ 1 + \frac{3}{4}\log_{e}(3) \right] – \sqrt{2} – \log_{e}(1+\sqrt{2}) – \frac{3}{2}\log_{e}(3)$$

$$2 + \frac{3}{2}\log_{e}(3) – \sqrt{2} – \log_{e}(1+\sqrt{2}) – \frac{3}{2}\log_{e}(3)$$

$$2 – \sqrt{2} – \log_{e}(1+\sqrt{2})$$

Ans. (2)