Answer :

Sure, let's evaluate the integral:

[tex]\[ \int \left( x + \frac{1}{x} \right) \, dx \][/tex]

We'll break the integral into two separate integrals:

[tex]\[ \int \left( x + \frac{1}{x} \right) \, dx = \int x \, dx + \int \frac{1}{x} \, dx \][/tex]

Now, let's evaluate each integral separately.

First, consider the integral:

[tex]\[ \int x \, dx \][/tex]

The integral of [tex]\( x \)[/tex] with respect to [tex]\( x \)[/tex] is:

[tex]\[ \int x \, dx = \frac{x^2}{2} + C_1 \][/tex]

where [tex]\( C_1 \)[/tex] is a constant of integration.

Next, consider the integral:

[tex]\[ \int \frac{1}{x} \, dx \][/tex]

The integral of [tex]\( \frac{1}{x} \)[/tex] with respect to [tex]\( x \)[/tex] is:

[tex]\[ \int \frac{1}{x} \, dx = \ln|x| + C_2 \][/tex]

where [tex]\( C_2 \)[/tex] is another constant of integration.

Putting these results together, we get:

[tex]\[ \int \left( x + \frac{1}{x} \right) \, dx = \frac{x^2}{2} + \ln|x| + C_1 + C_2 \][/tex]

Since [tex]\( C_1 \)[/tex] and [tex]\( C_2 \)[/tex] are both constants of integration, we can combine them into a single constant [tex]\( C \)[/tex].

Thus, the final answer is:

[tex]\[ \int \left( x + \frac{1}{x} \right) \, dx = \frac{x^2}{2} + \ln|x| + C \][/tex]