# How do you find the anti-derivative of 3 / x^3?

Mar 20, 2018

$\setminus \int \setminus \frac{3}{{x}^{3}} \mathrm{dx}$

First take out the constant "3"

$3 \setminus \int \setminus \frac{1}{{x}^{3}} \mathrm{dx}$

Apply property $\setminus \frac{1}{{x}^{3}} = {x}^{- 3}$

$3 \setminus \int \setminus {x}^{- 3} \mathrm{dx}$

$3 \setminus \frac{{x}^{- 3 + 1}}{- 3 + 1}$ =

$3 \left(\frac{{x}^{- 2}}{- 2}\right)$=

$3 \left(- \setminus \frac{1}{2 {x}^{2}}\right)$=

$- \setminus \frac{3}{2 {x}^{2}} + C$

Mar 20, 2018

$\frac{1.5}{{x}^{2}}$

#### Explanation:

Well, the antiderivative of a function is the same thing as the integral of the function.

So, the antiderivative of $\frac{3}{x} ^ 3$ is the same as $\int \frac{3}{x} ^ 3 \setminus \mathrm{dx}$.

And here we go,

$\int \frac{3}{x} ^ 3 \setminus \mathrm{dx}$

We take the constant out, and we get,

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

Using a little bit of algebraic manipulation, we get

$= 3 \int {\left({x}^{3}\right)}^{-} 1 \setminus \mathrm{dx}$

$= 3 \int {x}^{-} 3 \setminus \mathrm{dx}$

Now, we use the power rule, which states that $\int {x}^{n} \setminus \mathrm{dx} = \frac{{x}^{n + 1}}{n + 1}$, and so we get

$= 3 \cdot \frac{{x}^{- 3 + 1}}{- 3 + 1}$

$= 3 \cdot \frac{{x}^{-} 2}{-} 2$

$= \frac{- 3 {x}^{-} 2}{-} 2$

We also know that ${x}^{-} a = \frac{1}{{x}^{a}}$.

$= \frac{- 3}{- 2 {x}^{2}}$

$= \frac{1.5}{{x}^{2}}$