Lời giải:
\(I=\int ^{1}_{-1}\ln (x+\sqrt{1+x^2})dx\)
Chuyển $x\to -x$ thì:
\(I=\int ^{-1}_{1}\ln (-x+\sqrt{1+x^2})d(-x)\)
\(=-\int ^{-1}_{1}\ln (-x+\sqrt{1+x^2})dx=\int ^{1}_{-1}\ln (-x+\sqrt{1+x^2})dx\)
\(2I=\int ^{1}_{-1}[\ln (x+\sqrt{1+x^2})+\ln (-x+\sqrt{1+x^2})]dx\)
\(=\int^{1}_{-1}\ln [(x^2+1)-x^2]dx=\int^{1}_{-1}\ln 1dx=\int^{1}_{-1}0dx=0\)
$\Rightarrow I=0$