The integral $\int\left(1+x-\frac{1}{x}\right) e^{x+\frac{1}{x}} d x$ is equal to

$x e^{x+\frac{1}{x}}+C$

We have,

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

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

$=\int e^{x+\frac{1}{x}} d x+x e^{x+\frac{1}{x}}-\int e^{x+\frac{1}{x}} d x+C=x e^{x+\frac{1}{x}}+C$