The value of the integral \\int_{-1} ^{1} \log _{e}(\\sqrt{1-x}+\\sqrt{1+x}) d x is equal to : [20 Jul 2021 Shift 1]

Correct Answer is : 2loge2+π4−1

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

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

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

Definite Integrals Part 2

Section: Mathematics

Question : 81 of 100

Marks: +1, -0

The value of the integral

∫

−1

loge(√1−x+√1+x)dx is equal to :

[20 Jul 2021 Shift 1]

Go to Question:

Prev Question

Next Question