Let's find the indefinite integral of [tex]\(\frac{1}{x^2}\)[/tex] with respect to [tex]\(x\)[/tex].
We start with the integral:
[tex]\[
\int \frac{1}{x^2} \, dx
\][/tex]
We can rewrite the integrand in a form that might make it easier to integrate:
[tex]\[
\frac{1}{x^2} = x^{-2}
\][/tex]
Now, our integral becomes:
[tex]\[
\int x^{-2} \, dx
\][/tex]
To integrate [tex]\(x^{-2}\)[/tex], we use the power rule of integration which states that for any real number [tex]\(n \neq -1\)[/tex],
[tex]\[
\int x^n \, dx = \frac{x^{n+1}}{n+1} + C
\][/tex]
In our case, [tex]\(n = -2\)[/tex]. Applying the power rule:
[tex]\[
\int x^{-2} \, dx = \frac{x^{-2+1}}{-2+1} + C = \frac{x^{-1}}{-1} + C
\][/tex]
Simplifying the expression:
[tex]\[
\frac{x^{-1}}{-1} = -x^{-1}
\][/tex]
Recall that [tex]\(x^{-1}\)[/tex] is the same as [tex]\(\frac{1}{x}\)[/tex], so:
[tex]\[
-x^{-1} = -\frac{1}{x}
\][/tex]
Therefore, the indefinite integral of [tex]\(\frac{1}{x^2}\)[/tex] is:
[tex]\[
\int \frac{1}{x^2} \, dx = -\frac{1}{x} + C
\][/tex]
Here, [tex]\(C\)[/tex] is the constant of integration. This gives us the final answer:
[tex]\[
- \frac{1}{x} + C
\][/tex]
