Find f f' (x) = 1+3sqrt (x) f (4) = 25

It appears that you have a derivative and want to integrate ...

dy/dx=1+3√x

y (x) = x + (2/3) x^ (3/2) + C you are given that f (4) = 25 so we can solve for the constant of integration ...

y (4) = 25=4+16/3+C

21=16/3+C

(63-16) / 3=C

47/3=C so

f (x) = x + (2/3) x^ (3/2) + 47/3

f (x) = (3x+2x^ (3/2) + 47) / 3