Answer :

To solve the given integral
[tex]\[ \int \left(x + \frac{1}{x} \int \frac{1}{x} \, dx \right) \, dx \][/tex]
we need to go through it step by step.

1. Compute the inner integral:
First, we need to evaluate the inner integral
[tex]\[ \int \frac{1}{x} \, dx. \][/tex]
The integral of [tex]\( \frac{1}{x} \)[/tex] with respect to [tex]\( x \)[/tex] is simply [tex]\( \log(x) \)[/tex]. Therefore,
[tex]\[ \int \frac{1}{x} \, dx = \log(x). \][/tex]

2. Simplify the expression inside the outer integral:
Substitute the result of the inner integral back into the expression:
[tex]\[ x + \frac{1}{x} \log(x). \][/tex]

3. Compute the outer integral:
Now we need to evaluate the integral of the expression:
[tex]\[ \int \left( x + \frac{1}{x} \log(x) \right) \, dx. \][/tex]

This requires us to split the integral into two parts:
[tex]\[ \int x \, dx + \int \frac{\log(x)}{x} \, dx. \][/tex]

First, evaluate the integral of [tex]\(x\)[/tex]:
[tex]\[ \int x \, dx = \frac{x^2}{2}. \][/tex]

Next, evaluate the integral of [tex]\(\frac{\log(x)}{x}\)[/tex]:
[tex]\[ \int \frac{\log(x)}{x} \, dx. \][/tex]
This integral is a standard integral, and its result is known to be
[tex]\[ \int \frac{\log(x)}{x} \, dx = \frac{(\log(x))^2}{2}. \][/tex]

Combining these results, the integral of the entire expression is
[tex]\[ \int \left( x + \frac{1}{x} \log(x) \right) \, dx = \frac{x^2}{2} + \frac{(\log(x))^2}{2}. \][/tex]

So, the integral evaluates to:
[tex]\[ \int \left( x + \frac{1}{x} \log(x) \right) \, dx = \frac{x^2}{2} + x \log(x) - x. \][/tex]