Answer
$\text{(a) }2\\\text{(b) }0$
Work Step by Step
\[\begin{gathered}
part\,\,a \hfill \\
\int_{ - \infty }^\infty {{e^{ - \left| x \right|}}} dx \hfill \\
\hfill \\
the\,\,{\text{ }}function\,\,{e^{ - \left| x \right|}}is\,\,even,{\text{ then}} \hfill \\
\hfill \\
\int_{ - \infty }^\infty {{e^{ - \left| x \right|}}} dx = \hfill \\
\hfill \\
can\,\,be\,\,\,write\,\,as \hfill \\
\hfill \\
2\int_0^\infty {{e^{ - x}}} dx = 2 \hfill \\
\hfill \\
part\,b \hfill \\
\hfill \\
\int_{ - \infty }^\infty {\frac{{{x^3}}}{{1 + {x^8}}}\,} dx \hfill \\
\hfill \\
{\text{apply}}\,\,\,{\text{the}}\,\,{\text{properties}}\,\,{\text{of}}\,\,{\text{the}}\,\,{\text{integral}} \hfill \\
\hfill \\
\int_{ - \infty }^\infty {\frac{{{x^3}}}{{1 + {x^8}}}\,} dx = \int_0^\infty {\frac{{{x^3}}}{{1 + {x^8}}}\,} dx + \int_{ - \infty }^0 {\frac{{{x^3}}}{{1 + {x^8}}}\,} dx \hfill \\
\hfill \\
the\,\,fuction\,\,is\,\,odd,\,\,then \hfill \\
\hfill \\
= \int_0^\infty {\frac{{{x^3}}}{{1 + {x^8}}}\,} dx - \int_0^\infty {\frac{{{x^3}}}{{1 + {x^8}}}\,} dx = 0 \hfill \\
\end{gathered} \]
