First slide

Evaluation of definite integrals

Question

$\int_{- 1}^{1} \log (x + \sqrt{x^{2} + 1}) dx$ is equal to

Easy

A

0
B

log 2
C

$\log \frac{1}{2}$
D

None of these

Solution

Let $f (x) = \log (x + \sqrt{1 + x^{2}})$ and replacing x by we get
$\begin{array}{l} f (- x) = \log (\sqrt{1 + x^{2}} - x) \\ = \log (\sqrt{1 + x^{2}} - x) \frac{(\sqrt{1 + x^{2}} + x)}{(\sqrt{1 + x^{2}} + x)} \\ = \log \frac{[(1 + x^{2}) - x^{2}]}{(\sqrt{1 + x^{2}} + x)} \\ = \log 1 - \log (\sqrt{1 + x^{2}} + x) = - \log (\sqrt{1 + x^{2}} + x) \\ \Rightarrow f (- x) = - f (x) \end{array}$
Hence , f (x) is an odd function.
$∴ \int_{- 1}^{1} \log (x + \sqrt{1 + x^{2}}) dx = 0$

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