$4 \int_{0}^{1}\left(\frac{1}{\sqrt{3+x^{2}}+\sqrt{1+x^{2}}}\right) dx-3 \log _{e}(\sqrt{3})$ is equal to :

(B) $2-\sqrt{2}-\log _{e}(1+\sqrt{2})$

(A) $2+\sqrt{2}+\log _{e}(1+\sqrt{2})$

(C) $2+\sqrt{2}-\log _{e}(1+\sqrt{2})$

(D) $2-\sqrt{2}+\log _{e}(1+\sqrt{2})$

JEE Main · 2025 · 02 Apr 2025 · Shift 2 · Definite Integrals

4 \int_{0}^{1} (\frac{1}{3 + x ^{2} + 1 + x ^{2}}) d x - 3 lo g_{e} (3)

is equal to :

Answer: (B)

Rationalizing,

\frac{1}{3 + x ^{2} + 1 + x ^{2}} = \frac{3 + x ^{2} - 1 + x ^{2}}{2}

, so the expression becomes

2 \int_{0}^{1} (3 + x^{2} - 1 + x^{2}) d x - 3 ln 3

. Using

\int a^{2} + x^{2} d x = \frac{x}{2} a^{2} + x^{2} + \frac{a ^{2}}{2} ln (x + a^{2} + x^{2})

\int_{0}^{1} 3 + x^{2} d x = 1 + \frac{3}{4} ln 3

and

\int_{0}^{1} 1 + x^{2} d x = \frac{2}{2} + \frac{1}{2} ln (1 + 2)

. Hence the expression is

2 (1 + \frac{3}{4} ln 3 - \frac{2}{2} - \frac{1}{2} ln (1 + 2)) - \frac{3}{2} ln 3 = 2 - 2 - ln (1 + 2)

JEE Main · 2025 · 02 Apr 2025 · Shift 2 · Single correct option