Verify if f'(x)=g(x) . f(x)=x^3+ln(x), g(x)=1/x+3x^2.

We'll calculate the first derivative of
f(x).

f'(x) = (x^3+lnx)'

f'(x)
= 3x^2 + 1/x

The result of differentiating the function
f(x) represents the expression of the function g(x) = 3x^2 +
1/x.

We could also prove that f(x) is the antiderivative of
g(x) is calculating the indefinite integral of g(x), we'll get the expression of
f(x).

Int g(x)dx = f(x) +
C

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

We'll apply the property of the integral to be
additive:

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

 Int (3x^2)dx = 3Int
x^2dx

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

We'll simplify and we'll
get:

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

Int dx/x = ln x + C
(2)

Int g(x)dx = (1) +
(2)

Int g(x)dx = x^3 + ln x + C = f(x) + 0 =
f(x)

We notice that the result of the
indefinite integral of g(x) is the function f(x) =  x^3 + ln x (where C =
0).