To find the integral [tex]\(\int \frac{x^2 + 1}{x^2} \, dx\)[/tex], let's start by simplifying the integrand.
1. Simplify the integrand:
[tex]\[
\frac{x^2 + 1}{x^2}
\][/tex]
We can split the fraction into two separate terms:
[tex]\[
\frac{x^2}{x^2} + \frac{1}{x^2}
\][/tex]
2. Simplify each term:
[tex]\[
\frac{x^2}{x^2} = 1
\][/tex]
[tex]\[
\frac{1}{x^2} = x^{-2}
\][/tex]
So, the integrand becomes:
[tex]\[
1 + x^{-2}
\][/tex]
3. Integrate each term separately:
- The integral of [tex]\(1\)[/tex] with respect to [tex]\(x\)[/tex] is:
[tex]\[
\int 1 \, dx = x
\][/tex]
- The integral of [tex]\(x^{-2}\)[/tex] with respect to [tex]\(x\)[/tex] is:
[tex]\[
\int x^{-2} \, dx
\][/tex]
Recall the power rule for integration, [tex]\(\int x^n \, dx = \frac{x^{n+1}}{n+1} + C\)[/tex], for [tex]\(n \neq -1\)[/tex]. In this case, [tex]\(n = -2\)[/tex]:
[tex]\[
\int x^{-2} \, dx = \int x^{-2} \, dx = \frac{x^{-2+1}}{-2+1} = \frac{x^{-1}}{-1} = -x^{-1} = -\frac{1}{x}
\][/tex]
4. Combine the results:
[tex]\[
\int \frac{x^2 + 1}{x^2} \, dx = \int (1 + x^{-2}) \, dx = x - \frac{1}{x}
\][/tex]
So, the final answer for the integral [tex]\(\int \frac{x^2 + 1}{x^2} \, dx\)[/tex] is:
[tex]\[
x - \frac{1}{x} + C
\][/tex]
where [tex]\(C\)[/tex] is the constant of integration.
