The value of the integral $\int_{-1}^{1} \log _{e}(\sqrt{1-x}+\sqrt{1+x}) d x$ is equal to :

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

(A) $2 \log _{e} 2+\frac{\pi}{4}-1$

(C) $\frac{1}{2} \log _{\mathrm{e}} 2+\frac{\pi}{4}-\frac{3}{2}$

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

JEE Main · 2021 · 20 Jul 2021 · Shift 1 · Definite Integrals

The value of the integral

\int_{- 1}^{1} lo g_{e} (1 - x + 1 + x) d x

is equal to :

Answer: (B)

Squaring the sum gives

(1 - x + 1 + x)^{2} = 2 + 2 1 - x^{2}

, so the integrand splits as

lo g_{e} (1 - x + 1 + x) = \frac{1}{2} lo g_{e} 2 + \frac{1}{2} lo g_{e} (1 + 1 - x^{2})

The constant part contributes

lo g_{e} 2

over the interval of length

2

, leaving

lo g_{e} 2 + \frac{1}{2} K

with

K = \int_{- 1}^{1} lo g_{e} (1 + 1 - x^{2}) d x

. Integrating

K

by parts, the boundary term

[x lo g_{e} (1 + 1 - x^{2})]_{- 1}^{1}

vanishes because the logarithm is zero at both ends, and with

s = 1 - x^{2}

the surviving integrand simplifies through

\frac{x ^{2}}{s ( 1 + s )} = \frac{1 - s ^{2}}{s ( 1 + s )} = \frac{1}{s} - 1

K = \int_{- 1}^{1} \frac{x ^{2} d x}{1 - x ^{2} ( 1 + 1 - x ^{2} )} = \int_{- 1}^{1} (\frac{1}{1 - x ^{2}} - 1) d x = π - 2

The integral therefore equals

lo g_{e} 2 + \frac{π}{2} - 1

JEE Main · 2021 · 20 Jul 2021 · Shift 1 · Single correct option