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.
