# How do you integrate (1/x^4) dx?

Mar 23, 2018

$\int \frac{1}{x} ^ 4 \textcolor{w h i t e}{.} \mathrm{dx} = - \frac{1}{3 {x}^{3}} + C$

#### Explanation:

Note that:

$\frac{d}{\mathrm{dx}} \frac{1}{x} ^ 3 = \frac{d}{\mathrm{dx}} {x}^{- 3} = - 3 {x}^{- 4} = - 3 \left(\frac{1}{x} ^ 4\right)$

So:

$\int \frac{1}{x} ^ 4 \textcolor{w h i t e}{.} \mathrm{dx} = - \frac{1}{3 {x}^{3}} + C$

Mar 23, 2018

$\int \frac{1}{x} ^ 4 \mathrm{dx} = - \frac{1}{3} {x}^{-} 3 + C$

#### Explanation:

Recall that $\frac{1}{x} ^ a = {x}^{-} a$. We can then rewrite our integral as

$\int \frac{\mathrm{dx}}{x} ^ 4 = \int {x}^{-} 4 \mathrm{dx}$

Now, recall that $\int {x}^{a} \mathrm{dx}$ where $a \ne - 1$ is equal to ${x}^{a + 1} / \left(a + 1\right) + C$ where $C$ is the constant of integration. Then,

$\int {x}^{-} 4 \mathrm{dx} = {x}^{- 4 + 1} / \left(- 4 + 1\right) + C = - \frac{1}{3} {x}^{-} 3 + C = - \frac{1}{3 {x}^{3}} + C$