If f'(x) = 3x^2-5x+3

then f(x) =I = Int {3x^2-5x+3}dx

I = Int(3x^3) - Int(5x)dx) +Int (3x)dx

I = 3* (1/3+1)x^4-5(1/2) 5x^2 +3x+C , as Int x^n = (1/n+1)x^(n+1).

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

We know that F'(x) = f(x), where Int f'(x) dx = f(x).

We'll apply the indefinite integral to the expression of f(x):

Int f'(x) dx = Int (3x^2 - 5x + 3)dx

We'll apply the additive property of indefinite integrals:

Int (3x^2 - 5x + 3)dx = Int 3x^2dx - Int 5xdx + Int 3dx

Int 3x^2dx = 3x^3/3 + C

We'll simplify and we'll get:

Int 3x^2dx = x^3 + C (1)

Int 5xdx = 5x^2/2 + C (2)

Int 3dx = 3x + C (3)

We'll add (1),(2),(3):

f(x) = (1)+(2)+(3)

**f(x) = x^3 + 5x^2/2 + 3x + C**

Note: C+C+C = C (family of constants)

Here we are given f'(x) = 3x^2 - 5x + 3. If we integrate f'(x) we get f(x).

Now the integral of x^n is x^(n+1)/ (n+1)

Int [ 3x^2 - 5x + 3 ] = Int [ 3x^2] - Int [ 5x] + Int 3

=> 3x^3/3 - 5x^2 / 2 + 3x

=> 3x^3 - (5/2) x^2 + 3x

**Therefore f(x) = 3x^3 - (5/2) x^2 + 3x + C**