# What is the antiderivative of 1 / (x^2)?

$- \frac{1}{x} + c$
For this antiderivative, you would use the power rule for antiderivatives/integrals. This states that $\int {x}^{n} = \frac{1}{n + 1} \left({x}^{n + 1}\right)$. Since $\frac{1}{x} ^ 2 = {x}^{-} 2$ and $n \ne - 1$ in this case, you can apply this power rule.
Answer: $\int \frac{1}{x} ^ 2 = \int {x}^{-} 2 = {x}^{-} \frac{1}{-} 1 + c = - \frac{1}{x} + c$