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

We begin with the integral

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

A very common strategy for such problems is to look for a function whose derivative reproduces the given integrand. Because the exponential factor $$e^{\,x+\frac{1}{x}}$$ is present everywhere, we suspect a product of this exponential with some elementary function of $$x$$. Let us therefore examine the derivative of the product $$x\,e^{\,x+\frac{1}{x}}.$$

First we recall the product rule of differentiation:

$$\frac{d}{dx}\,[u(x)\,v(x)] \;=\; u'(x)\,v(x) \;+\; u(x)\,v'(x).$$

Here we choose

$$u(x)=x, \qquad v(x)=e^{\,x+\frac{1}{x}}.$$

We compute each derivative separately.

Clearly,

$$u'(x)=\frac{d}{dx}(x)=1.$$

Next, to differentiate $$v(x)=e^{\,x+\frac{1}{x}},$$ we first note the chain rule:

$$\frac{d}{dx}\,e^{f(x)} = e^{f(x)}\;f'(x).$$

In our case $$f(x)=x+\frac{1}{x}.$$ Therefore

$$f'(x)=\frac{d}{dx}\Bigl(x+\frac{1}{x}\Bigr)=1-\frac{1}{x^{2}}.$$

Applying the chain rule gives

$$v'(x)=e^{\,x+\frac{1}{x}}\left(1-\frac{1}{x^{2}}\right).$$

Now we use the product rule:

$$\frac{d}{dx}\left[x\,e^{\,x+\frac{1}{x}}\right] \;=\; u'(x)v(x) + u(x)v'(x).$$

Substituting the values we have just obtained,

$$\frac{d}{dx}\left[x\,e^{\,x+\frac{1}{x}}\right] \;=\; 1\cdot e^{\,x+\frac{1}{x}} \;+\; x\;e^{\,x+\frac{1}{x}}\left(1-\frac{1}{x^{2}}\right).$$

We now factor out the common exponential:

$$\frac{d}{dx}\left[x\,e^{\,x+\frac{1}{x}}\right] \;=\; e^{\,x+\frac{1}{x}}\Bigl[\,1 + x\Bigl(1-\frac{1}{x^{2}}\Bigr)\Bigr].$$

Inside the square brackets we simplify the expression:

$$x\Bigl(1-\frac{1}{x^{2}}\Bigr) \;=\; x\cdot 1 - x\cdot\frac{1}{x^{2}} \;=\; x - \frac{1}{x}.$$

Hence

$$\frac{d}{dx}\left[x\,e^{\,x+\frac{1}{x}}\right] \;=\; e^{\,x+\frac{1}{x}}\Bigl(1 + x - \frac{1}{x}\Bigr).$$

Observe that the factor in parentheses is exactly the same as the factor in the integrand. Therefore,

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

When an integrand is recognized as a total derivative, the integration becomes immediate:

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

where $$C$$ is the constant of integration.

On comparing with the given options, this result corresponds to

$$xe^{\,x+\frac{1}{x}} + c,$$

which is Option D.

Hence, the correct answer is Option D.