∫−aa log⁡x+x2+1dx=

I=∫−aa log⁡x+x2+1dx=∫−aa log⁡−x+x2+1dx∫−aa f(x)dx=∫−aa f(−x)dx∫−aa log⁡−x2+x2+1x+x2+1dx=−∫−aa log⁡x+x2+1dx=−I⇒I=0

∫−aa log⁡x+x2+1dx=

I=∫−aa log⁡x+x2+1dx=∫−aa log⁡−x+x2+1dx∫−aa f(x)dx=∫−aa f(−x)dx∫−aa log⁡−x2+x2+1x+x2+1dx=−∫−aa log⁡x+x2+1dx=−I⇒I=0

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$\int_{- a}^{a} \log (x + \sqrt{x^{2} + 1}) d x =$

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$2 \log (a^{2} + 1)$

$2 \log (\sqrt{a^{2} + 1} - a)$

$2 \log (a + \sqrt{a^{2} + 1})$

answer is C.

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Detailed Solution

$\begin{array}{l} I = \int_{- a}^{a} \log (x + \sqrt{x^{2} + 1}) d x = \int_{- a}^{a} \log (- x + \sqrt{x^{2} + 1}) d x \\ (\int_{- a}^{a} f (x) d x = \int_{- a}^{a} f (- x) d x) \\ \int_{- a}^{a} \log (\frac{- x^{2} + x^{2} + 1}{x + \sqrt{x^{2} + 1}}) d x = - \int_{- a}^{a} \log (x + \sqrt{x^{2} + 1}) d x \\ = - I \\ \Rightarrow I = 0 \end{array}$