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

$(x+1) e^{x+\frac{1}{x}}+C$ $-x e^{x+\frac{1}{x}}+C$ $(x-1) e^{x+\frac{1}{x}}+C$ $x e^{x+\frac{1}{x}}+C$

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

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