🛠️ JEE➗ Maths

The value of the integral \({\int }_{-1}^{1}{\log }_{e}(\sqrt{1-x}+\sqrt{1+x})dx\) is equal to : [JEE Main 2021, 20 Jul …

Q1

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

[JEE Main 2021, 20 Jul (Shift 1)]

a

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

b

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

c

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

d

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

🔒

Answer & explanation — PYQ Pass

Options are free to see. Unlock the correct answer and full explanation with Pass.

Unlock · ₹149

← Previous question Next question →

Practice more JEE Maths PYQs

See every question on Definite Integration, or browse the full JEE question bank.

See all questions on Definite Integration →