# Show that F(x)=1/3x-5/9ln(3x+5) is primitive of f(x)=x/3x+5?

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### 1 Answer

You have two alternatives, either differentiate F(x) with respect to x, to check if it gives f(x), or you may consider to integrate f(x) to check if the result is F(x).

Using the first alternative, you should differentiating F(x) with respect to x such that:

`F'(x) = ((1/3)x - (5/9)ln(3x+5))'`

`F'(x) = 1/3 - (5/9)*(1/(3x+5))*(3x+5)'`

`F'(x) = 1/3 - (5/9)*(3/(3x+5)) => F'(x) = 1/3 - (5/(3(3x+5)))`

You need to bring the fractions to a common denominator such that:

`F'(x) = (3x + 5 - 5)/(3(3x+5)) => F'(x) = (3x)/(3(3x+5)) `

`F'(x) = x/(3x+5)`

**Notice that `F'(x) = f(x) = x/(3x+5)` , hence, F(x) is the primitive of f(x).**

Using the second alternative, you should integrate f(x) such that:

`int x/(3x + 5)dx = (1/3)int (3x)/(3x+5)dx`

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

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

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

`int x/(3x + 5)dx = (1/3)x - (5/3)*(1/3)*ln|3x+5| + c`

`int x/(3x + 5)dx = (1/3)x - (5/9)*ln|3x+5| + c`

**Notice that integrating f(x) yields F(x), hence F(x) is the primitive of f(x).**

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