\(I =\int _{\frac{-\pi }{2} }^{\frac{\pi }{2} }\ln (x+\sqrt{x^{2} +1} ) \, dx \)
Đặt \(x=-t\Rightarrow dx=-dt,\) khi đó :
\(I =\int _{\frac{-\pi }{2} }^{\frac{\pi }{2} }\ln (x+\sqrt{x^{2} +1} ) \, dx\)
\(=-\int _{\frac{\pi }{2} }^{\frac{-\pi }{2} }\ln (-x+\sqrt{x^{2} +1} \, ) \, dx \)
\(=\int _{\frac{-\pi }{2} }^{\frac{\pi }{2} }\ln (-x+\sqrt{x^{2} +1} \, ) \, dx=\int _{\frac{-\pi }{2} }^{\frac{\pi }{2} }\ln \frac{1}{x+\sqrt{x^{2} +1} } \, dx
\)
\(=-\int _{\frac{-\pi }{2} }^{\frac{\pi }{2} }\ln (x+\sqrt{x^{2} +1} ) \, dx=-I . \)
Suy ra : \(2I =0\Rightarrow I =0\)
Đáp số : \( I =0\).