f'(x) = (2x-4)/(3x+5)

f(x) = integral f'(x)

= intg (2x-4)/(3x+5) dx

Let t= 3x + 5 ==> x = (t-5)/3

=> dt = 3 dx

==> f(x) = intg (2(t-5)/3 - 4 ]/t

= intg ((2t - 10 - 12)/3t dt/3

= intg (2t-22)/9t dt

= intg 2/9 - 22/9t dt

= (2/9)t - 22/9 ln t + C

** ==> f(x) = (2/9)(3x+5) + (22/9)*ln (3x+5) + C**

To determine the function f(x), we'll have to integrate (2x-4)/(3x+5)

Int (2x-4)dx/(3x+5) = f(x) + C

To determine the integral of f'(x), we'll use substitution technique, by changing the variable x.

We'll note 3x + 5 = t

We'll subtract 5 both sides:

3x = t-5

We'll divide by 3:

x = (t-5)/3

We'll differentiate both sides:

dx = dt/3

We'll substitute in original integral:

Int (2x-4)dx/(3x+5) = Int [2((t-5)/3)-4]dt/3t

(2/3)*Int [(t-5)/3 - 2]dt/t = (2/3)*Int (t-5-6)dt/3t

(2/9)*Int (t-11)dt/t

We'll use the additive property of the integral:

(2/9)*Int (t-11)dt/t = (2/9)*Int tdt/t - (2/9)*Int 11dt/t

(2/9)*Int (t-11)dt/t = (2/9)*Int dt - 22/9 Int dt/t

(2/9)*Int (t-11)dt/t = 2t/9 - (22/9)*ln |t| + C

**Int (2x-4)dx/(3x+5) = 2(3x+5)/9 - (22/9)*ln|3x+5| + C**

f'(x) = (2x-4)/(3x+5).

To find f(x) we have to integrate (2x-4)/(3x+5).

2x-4 = 2/3(3x-5) + 10/3 -4.

Therefore (2x-4)/(3x+5) = 1/2 - (2/3)/(3x+5)

Therefore Int (2x-4)/(3x+5) = Int(2/3)dx -(2/3)Int dx/(3x+5) = 2x/3 +(2/3)(1/3) log(3x+5) + const

Therfore f(x) = Int f'(x) dx = Int {(2x-4)/(3x+5)}dx = 2x/3 + (2/9) ln(3x+5) + C