# How do you find the antiderivative of 2/z^2?

Apr 30, 2017

Represent the fraction as a negative exponent and use the reverse power rule.

#### Explanation:

$\frac{2}{z} ^ 2$ can be rewritten as $2 {z}^{-} 2$
So our integral becomes:
$\int 2 {z}^{-} 2 \mathrm{dz}$
we can now pull out the $2$ from the integral as it is a constant.
$2 \int {z}^{-} 2 \mathrm{dz}$
The reverse power rule states $\int {x}^{n} \mathrm{dx} = {x}^{n + 1} / \left(n + 1\right) + c$
Thus $2 \int {z}^{-} 2 \mathrm{dz} = 2 \cdot \left[{z}^{- 2 + 1} / \left(- 2 + 1\right)\right] = 2 \cdot \left[{z}^{-} \frac{1}{-} 1\right] = 2 \cdot \left[- \frac{1}{z}\right]$
Simplifying this we get: $- \frac{2}{z} + c$
We add a $+ c$ at the end because the integral is indefinite.