_ -1 ^1 (x^2+1+x ) d x= - Vidyadip

MCQ

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

✓
$0$
B
$\log 2$
C
$\log \frac{1}{2}$
D
1

✓

Answer

Correct option: A.

$0$

(A)
Let $f (x)=\log \left(\sqrt{1+x^2}+x\right)$
$\therefore \quad f (-x)=\log \left(\sqrt{1+x^2}+x\right)$
$=\log \left(\sqrt{1+x^2}-x\right) \cdot \frac{\left(\sqrt{1+x^2}+x\right)}{\left(\sqrt{1+x^2}+x\right)}$
$\begin{array}{l}=\log \left(\frac{1+x^2-x^2}{\sqrt{1+x^2}+x}\right) \\ =\log 1-\log \left(\sqrt{1+x^2}+x\right) \\ =-\log \left(\sqrt{1+x^2}+x\right)=- f (x)\end{array}$
$\therefore f (x)$ is an odd function.
$\therefore \quad \int_{-1}^1 \log \left(\sqrt{1+x^2}+x\right) d x=0$

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