The integral ∫ 1 + x - 1 x e x + 1 x d x , is equal to

$\int \left(1+x-\frac{1}{x}\right)e^{x+\frac{1}{x}}dx=\int e^{x+\frac{1}{x}}d{x}^{}+\int x\left(1-\frac{1}{x^2}\right)e^{x+\frac{1}{x}}dx$

Integrating $\int \underset{I}{\underbrace{x}}.\underset{II}{\underbrace{(1-\frac{1}{x^2})e^{x+\frac{1}{x}}}}dx$, by parts

$I=x{e}^{x+\frac{1}{x}}-\int e^{x+\frac{1}{x}}dx+c$

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

$=x{e}^{x+\frac{1}{x}}+c$, where $c$ is the constant of integration.