Only Partial Fools

Consider,

$\cfrac { \partial \,  }{ \partial \, x } \left\{ ln(\psi ) \right\} =\cfrac { 1 }{ \psi  } \cfrac { \partial \, \psi  }{ \partial \, x }$

Integrating,

$\int { \cfrac { \partial \,  }{ \partial \, x } \left\{ ln(\psi ) \right\}  } \, d\, x=\int { \cfrac { 1 }{ \psi  } \cfrac { \partial \, \psi  }{ \partial \, x }  } \, d\, x=\int { \cfrac { 1 }{ \psi  }  } \, d\, \psi =ln(\psi )$

So, in general,

$\int { \cfrac { \partial \,  }{ \partial \, x } \left\{ f(\psi ) \right\}  } \, d\, x=\int { f^{ ' }(\psi )\cfrac { \partial \, \psi  }{ \partial \, x }  } \, d\, x=\int { f^{ ' }(\psi ) } \, d\, \psi =f(\psi )$

As if,$\require{cancel}$
$\int { f^{ ' }(\psi )\cfrac { \partial \, \psi  }{ \cancel{\partial \, x} }  } \,\cancel{ d\, x}=\int { f^{ ' }(\psi ) } \, d\, \psi$

There can only be partial derivative, what you don't know does not made it full.