Question

Evaluate the integral
${\int }_{-3}^3\log (\sqrt{x^2+1}+x)dx=$

• A. $0$
• B. $\log 2$
• C. $-\log 2$
• D. $2\log 2$

Answer 1

The correct option is A $0$

$I={\int }_{-3}^{2}log(\sqrt{x^2+1}+x)dx$ ---(1)

${\int }_a^{b}f\left(x\right)dx={\int }_a^{b}f(a+b-x)dx$

$I={\int }_{-3}^{3}log(\sqrt{1+x^2}-x)dx$ ---(2)

Adding 1 and 2, we get

$2I={\int }_{-3}^{3}log\left(1\right)$

$2I=0$

$I=0$