The work done is the line integral of the sum of the forces acting on the object. $$W=\int\limits_{\ell=0}^{\ell=h}(\vec{F}_{\rm lift}+\vec{F}_g)\cdot d\vec{l} =mgh,$$ where $\vec{F}_g$ is in the opposite direction to $\vec{F}_{\rm lift}$.

The only way to can get the object to move upwards is if the magnitudes $F_{\rm lift} > F_g$.

However, if you want it to stop at $h$, then at some point in the second phase of the motion then the magnitudes $F_{\rm lift} < F_g$.

If you apply a constant lifting force, then of course the body will not be at rest when it reaches $h$. You will have done more work and the extra work will be accounted for by the kinetic energy of the body.

The work done is the line integral of the sum of the forces acting on the object. $$W=\int\limits_{\ell=0}^{\ell=h}(\vec{F}_{\rm lift}+\vec{F}_g)\cdot d\vec{l} =mgh,$$ where $\vec{F}_g$ is in the opposite direction to $\vec{F}_{\rm lift}$.

The only way to can get the object to move upwards is if the magnitudes $F_{\rm lift} > F_g$.

However, if you want it to stop at $h$, then at some point in the second phase of the motion then the magnitudes $F_{\rm lift} < F_g$.

If you apply a constant lifting force, then of course the body will not be at rest when it reaches $h$. You will have done more work and the extra work will be accounted for by the kinetic energy of the body.

I do not understand what you mean by "bring it to rest". That implies exerting a force in the opposite direction to gravity. i.e. reducing the magnitude of $F_{\rm lift}$ to a negative value, which will jusr reduce the time duration of the necessary second phase beyond that if the decelerating force is provided by gravity alone.

No, the work done will not be the same, it will be greater. Some of that work will go into providing the object with kinetic energy. How do you propose to bring it to rest? If it is by applying a force in the opposite direction then the sum of TWO line integrals will result in $mgh$ as you had before.

$$W=\int\limits_{\ell=0}^{\ell=h}\vec{F}_{\rm lift}\cdot\vec{d\ell} + \int \vec{F}_{\rm stop}\cdot \vec{d\ell} =mgh,$$ where $\vec{F}_{\rm stop}$ is your stopping force and will be in the opposite direction to $\vec{F}_{\rm lift}$.

The work done is the line integral of the sum of the forces acting on the object. $$W=\int\limits_{\ell=0}^{\ell=h}(\vec{F}_{\rm lift}+\vec{F}_g)\cdot d\vec{l} =mgh,$$ where $\vec{F}_g$ is in the opposite direction to $\vec{F}_{\rm lift}$.

The only way to can get the object to move upwards is if the magnitudes $F_{\rm lift} > F_g$.

However, if you want it to stop at $h$, then at some point in the second phase of the motion then the magnitudes $F_{\rm lift} < F_g$.

If you apply a constant lifting force, then of course the body will not be at rest when it reaches $h$. You will have done more work and the extra work will be accounted for by the kinetic energy of the body.