${\int }_{ }^{ }\left(1+x-x^{-1}\right)e^{x+x^{-1}}dx$ is equal to

$\int \left(1+x-x^{-1}\right)e^{x+x^{-1}}dx$

$=\int e^{x+x^{-1}}dx+\int \left(x-\frac{1}{x}\right)e^{x+x^{-1}}dx$

$=e^{x+x^{-1}}\int dx-\int \left[\frac{d}{dx}{e}^{x+x^{-1}}\right]\left(x\right)dx+\int \left(x-\frac{1}{x}\right)e^{x+x^{-1}}dx$

$=x{e}^{x+x^{-1}}-\underset{ }{\overset{ }{\int }}\left(x-\frac{1}{x}\right)e^{x+x^{-1}}dx+\underset{ }{\overset{ }{\int }}\left(x-\frac{1}{x}\right)e^{x+x^{-1}}dx+c$

$=x{e}^{x+x^{-1}}+C$