Yes. If you know the derivative of a function you can identify the original function by integrating what you have. During integration constants in the original function cannot be identified as they are eliminated during differentiation. But this is the only way to go about it.
Only integrating the function you can find the original function, or the primitive.
For instance, we know that F'(x) = f(x)
The inverse process is Int f'(x)dx = F(x).
If the function is f(x) = sqrtx /(1+sqrtx) => F(x) = Int f(x)dx
Int f(x)dx = Int sqrtx dx/(1+sqrtx)
We'll substitute 1 + sqrt x = t.
We'll differentiate both sides:
dx/2sqrt x = dt
dx = (2sqrt x)dt
But sqrt x = t - 1
dx = 2(t - 1)dt
We'll re-write the integral;
Int f(x)dx = Int 2(t - 1)^2dt/t
We'll expand the square;
Int 2(t - 1)^2dt/t = 2 Int (t^2 - 2t + 1)/t
We'll apply the property of integrals to be additive:
2 Int (t^2 - 2t + 1)/t = 2Int t^2dt/t - 4Int tdt/t + 2Int dt/t
2 Int (t^2 - 2t + 1)/t = 2Int tdt - 4 Int dt + 2Int dt/t
2 Int (t^2 - 2t + 1)/t = 2*t^2/2 - 4t + 2ln|t| + C
We'll simplify and we'll get:
2 Int (t^2 - 2t + 1)/t = t^2 - 4t + 2ln|t| + C
F(x) = (1 + sqrt x)^2 - 4(1 + sqrt x) + ln (1 + sqrt x)^2 + C