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  • The value of the integral  <img alt="\int_{-1}^{1} \log \left(x+\sqrt{x^{2}+1}\right) d x" src="https://entrancecorner.codecogs.com/gif.latex?%5Cint_%7B-1%7D%5E%7B1%7D%20%5Clog%20%5Cleft%28x&plus;%5Csqrt%7Bx%5E%7B2%7D&plus;1%7D%5Cright%29%20d

The value of the integral  \int_{-1}^{1} \log \left(x+\sqrt{x^{2}+1}\right) d x is :
Option: 1 2
Option: 2 2
Option: 3 2
Option: 4 2
Option: 5 0
Option: 6 0
Option: 7 0
Option: 8 0
Option: 9 -1
Option: 10 -1
Option: 11 -1
Option: 12 -1
Option: 13 1
Option: 14 1
Option: 15 1
Option: 16 1

Answers (1)

best_answer

Let\, f\left ( x \right )= \log \left ( x+\sqrt{x^{2}+1} \right )
So\, f\left (- x \right )= \log \left ( \sqrt{x^{2}+1}-x \right )                        = \log\left ( \frac{\left ( \sqrt{x^{2}+1}-x \right )\left ( \sqrt{x^{2}+1}+x \right )}{\left ( \sqrt{x^{2}+1}+x \right )} \right )
                      = \log\left ( \frac{1}{\sqrt{x^{2}+1}+x} \right )
                      = -\log \left ( \sqrt{x^{2}+1}+x\right )
                      = -f\left ( x \right )
so f(x) is an add function
\therefore \int_{-1}^{1}f\left ( x \right )dx= 0

Posted by

Kuldeep Maurya

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