Given the first derivative:

f'(x) = (2x+3)/x

We need to find f(x).

First we will simplify f'(x).

==> f'(x) = 2x/x + 3/x

==> f'(x) = 2 + 3/x

Now we know that f(x) = intg f'(x).

==> f(x) = intg ( 2 + 3/x) dx

= intg 2 + intg 3/x dx

= 2x + 3*lnx + C

==> f(x) = 2x + 3*lnx + C

But we are given that f(1) = 3

==> 2 + 3ln1 + C = 3

==> But ln 1 = 0

==> 2 + C = 3

==> C = 1

Then we conclude that:

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

f'(x) = 2x+3)/x. To find f(x) if f(1) = 3.

f'(x) = (2x+3)/x.

f'(x) = 2+3/x.

=> f(x) = Int f'(x) dx = Int 2dx + int (2/x) dx .

=> f(x) = 2x+3lnx +C...(1), where C is the constant of integration.

=> f(1) = 2*1+3ln1+C, as ln= 0 and given f(1) = 3.

=> 3 = 2+0+C.

=> C = 3-2 = 1. Substitute this value of C in (1):

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

So f(x) = 2x+3lnx +1.