Question

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

Answer 1

`\int \frac{3}{x^3}dx`

First take out the constant "3"

`3\int \frac{1}{x^3}dx`

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

`3\int \x^{-3}dx`

`3 \frac{x^{-3+1}}{-3+1}` =

`3(frac{x^{-2}}{-2})`=

`3(-\frac{1}{2x^2})`=

Final Answer:

`-\frac{3}{2x^2}+C`

Answer 2

`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 `3/x^3` is the same as `int3/x^3 \ dx`.

And here we go,

`int3/x^3 \ dx`

We take the constant out, and we get,

`=3int1/x^3 \ dx`

Using a little bit of algebraic manipulation, we get

`=3int(x^3)^-1 \ dx`

`=3intx^-3 \ dx`

Now, we use the power rule, which states that `intx^n \ dx=(x^(n+1))/(n+1)`, and so we get

`=3*(x^(-3+1))/(-3+1)`

`=3*(x^-2)/-2`

`=(-3x^-2)/-2`

We also know that `x^-a=1/(x^a)`.

`=(-3)/(-2x^2)`

`=1.5/(x^2)`