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]
[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]