Sure, let's solve the integral [tex]\(\int \frac{1}{x^2} \, dx\)[/tex] step-by-step.
1. Identify the form of the integrand.
The integrand here is [tex]\(\frac{1}{x^2}\)[/tex]. This can be rewritten using a negative exponent as [tex]\(x^{-2}\)[/tex].
2. Set up the integral.
[tex]\[
\int x^{-2} \, dx
\][/tex]
3. Apply the power rule for integration.
The power rule states that [tex]\(\int x^n \, dx = \frac{x^{n+1}}{n+1} + C\)[/tex] for [tex]\(n \neq -1\)[/tex].
Here, [tex]\(n = -2\)[/tex].
[tex]\[
\int x^{-2} \, dx = \frac{x^{-2+1}}{-2+1} + C = \frac{x^{-1}}{-1} + C
\][/tex]
4. Simplify the expression.
[tex]\[
\frac{x^{-1}}{-1} + C = -x^{-1} + C
\][/tex]
5. Rewrite the expression using positive exponents.
[tex]\[
-x^{-1} + C = -\frac{1}{x} + C
\][/tex]
Therefore, the integral [tex]\(\int \frac{1}{x^2} \, dx\)[/tex] is [tex]\(-\frac{1}{x} + C\)[/tex].
Now, examining the given options:
A) [tex]\(-x^2\)[/tex]
B) [tex]\(\frac{2}{x}\)[/tex]
C) [tex]\(\frac{1}{x^2}\)[/tex]
D) [tex]\(-\frac{1}{x} + C\)[/tex]
The correct answer is D) [tex]\(-\frac{1}{x} + C\)[/tex].
