# Find the primitive function of (x^2+x+3)/3x^5

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### 2 Answers

**The primitive function of a given function is the antiderivative or integral of that given function.**

`f(x) = (x^2+x+3)/(3x^5)`

I will rewrite this expression as below.

`f(x) = 1/3 ((x^2+x+3)/(x^5))`

and dividing both numerator and denominator by `x^5` .

`f(x) = 1/3(x^(-3)+x^(-4)+3x^(-5))`

To find the primitive function of f(x), I can integrate it wrt x.

If the primitive function of f(x) = g(x), then,

`g(x) = intf(x)dx`

`g(x) = int1/3(x^(-3)+x^(-4)+3x^(-5))dx`

`g(x) = 1/3int(x^(-3)+x^(-4)+3x^(-5))dx`

Integrating,

`g(x) = 1/3(x^(-2)/(-2)+x^(-3)/(-3)+(3x^-4)/(-4))+c`

This gives,

`g(x) = -1/3(1/(2x^2)+1/(3x^3)+3/(4x^4))`

Rearranging,

`g(x) = -1/3((6x^2+4x+9)/(12x^4))`

Therefore, `g(x)` is equal to,

`g(x) = -(6x^2+4x+9)/(36x^4)`

**The primitive function of** `f(x) = (x^2+x+3)/(3x^5)` is `g(x) = -(6x^2+4x+9)/(36x^4)`.

**Sources:**

The primitive function of `(x^2+x+3)/(3x^5)` has to be determined.

The primitive function of `(x^2+x+3)/(3x^5)` is `int (x^2+x+3)/(3x^5) dx`

`int (x^2+x+3)/(3x^5) dx`

=> `(1/3)*int (x^(-3) dx + (1/3)*int x^(-4) dx + int x^-5 dx`

=> `(-1/6)*x^(-2)+ (-1/9)x^(-3) dx + (-1/4)*x^-4`

=> `-(6x^2 + 4x + 9)/(36x^4)`

**The primitive function of `(x^2+x+3)/(3x^5)` is `-(6x^2 + 4x + 9)/(36x^4)`**