Answer: \(\mathrm{e}^{1+x^{2}}+\mathrm{e}^{1-x}=2\mathrm{e}-\mathrm{e}x+\frac{3}{2}\mathrm{e}x^{2}+\ldots\)

Use the Maclaurin expansion \(\mathrm{e}^u=1+u+\frac{u^2}{2}+\ldots\).

First,

\(\mathrm{e}^{1+x^2}=\mathrm{e}\,\mathrm{e}^{x^2}=\mathrm{e}\left(1+x^2+\frac{x^4}{2}+\ldots\right)\).

So, up to and including the term in \(x^2\),

\(\mathrm{e}^{1+x^2}=\mathrm{e}+\mathrm{e}x^2+\ldots\)

Next,

\(\mathrm{e}^{1-x}=\mathrm{e}\,\mathrm{e}^{-x}=\mathrm{e}\left(1-x+\frac{x^2}{2}+\ldots\right)\)

so

\(\mathrm{e}^{1-x}=\mathrm{e}-\mathrm{e}x+\frac{1}{2}\mathrm{e}x^2+\ldots\)

Adding the two series gives

\(\mathrm{e}^{1+x^2}+\mathrm{e}^{1-x}=(\mathrm{e}+\mathrm{e}x^2)+(\mathrm{e}-\mathrm{e}x+\frac{1}{2}\mathrm{e}x^2)+\ldots\)

\(=2\mathrm{e}-\mathrm{e}x+\frac{3}{2}\mathrm{e}x^2+\ldots\)

Hence the Maclaurin series up to and including the term in \(x^2\) is \(\boxed{2\mathrm{e}-\mathrm{e}x+\frac{3}{2}\mathrm{e}x^2}\).