02-02-2019, 12:18 AM
\\begin{array}{*{20}{c}}
0\\
b
\end{array}\\
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \mathop {\lim }\limits_{b \to - \infty } \left. {\left( {\sqrt 2 - \sqrt {{b^2} + 2} } \right)} \right] = \sqrt 2 - \infty = - \infty \,\,\\
2)In\,same\,way\,\,\,\,\int\limits_0^\infty {\frac{x}{{\sqrt {{x^2} + 2} }}} \,\,dx = \mathop {\lim }\limits_{a \to \infty } \int\limits_0^a {\frac{x}{{\sqrt {{x^2} + 2} }}} \,\,dx = \mathop {\lim }\limits_{a \to \infty } \left. {\sqrt {{x^2} + 2} } \right]\begin{array}{*{20}{c}}
a\\
0
\end{array}\\
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \mathop {\lim }\limits_{a \to \infty } \left. {\left( {\sqrt {{a^2} + 2} - \sqrt 2 } \right)} \right] = \infty - \sqrt 2 = \infty \,\,\\
Hence\,\,\int\limits_{ - \infty }^\infty {\frac{x}{{\sqrt {{x^2} + 2} }}} \,\,dx\,\,\,is\,diverges
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