# Prove that f is the antiderivative of g is f(x)=x^3+lnx and g(x)=3x^2+1/x

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To prove that anti derivative of g(x) = 3x^2+1/x is f(x) = x^3+lnx.

We start with f(x) = x^3+lnx.

We differentiate both sides:

f'(x) = d/dx{x^3+lnx}

f'(x) = d/dx(x^3)+d/dx(lnx)

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

Thus we got d/dx f(x) = d/dx{x^3+lnx} = 3x^2+1/x.

Therefore , the antiderivative of (3x^2+1/x) is x^3+lnx

If f(x) is the antiderivative of g(x) => f'(x) = g(x)

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

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

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

We notice that the result of differentiating the function f(x) is 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

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