# How do you integrate int 1/(3x)^2dx?

##### 1 Answer
Feb 7, 2017

$\int \frac{1}{3 x} ^ 2 \mathrm{dx} = - \frac{1}{9 x} + C$

#### Explanation:

Since ${\left(3 x\right)}^{2} = 3 x \cdot 3 x = 9 \cdot {x}^{2}$

We can say

$\int \frac{1}{3 x} ^ 2 \mathrm{dx} = \int \frac{1}{9 {x}^{2}} \mathrm{dx}$

and since $\int c f \left(x\right) \mathrm{dx} = c \int f \left(x\right) \mathrm{dx}$

$= \frac{1}{9} \int \frac{1}{x} ^ 2 \mathrm{dx}$

and since $\frac{1}{x} ^ 2 = {x}^{- 2}$

$= \frac{1}{9} \int {x}^{-} 2 \mathrm{dx}$

Since the Fundamental Theorem of Calculus says

$\int f ' \left(x\right) \mathrm{dx} = f \left(x\right) + C$

and

$\left(- {x}^{- 1}\right) ' = - \left(- {x}^{- 2}\right) = {x}^{- 2}$

by the power rule

$= - \frac{1}{9} {x}^{-} 1 + C = - \frac{1}{9 x} + C$