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regex - how to remove all characters from beginning of line, up to, and including the first space/whitespace character

today i had some console output from an error i got while debugging some code ("..." means that i've shortened the actual text bc it was so long):
12:12:54,828 INFO  [com.myapp.tiv.persistentservices.transportagreement.daoimpl....
12:13:02,216 INFO  [stdout] (http-/ setErrorMessage: Feil oppstod...
12:13:02,217 ERROR [stderr] (http-/ org.hibernate.LazyInitializat...
12:13:02,218 ERROR [stderr] (http-/     at org.hibernate.proxy.Abstr...

so the red text is what i wanted to remove so i could use the rest of the output in an explanation in our bugtracker. to do this, i created the following regex search/replace code that i ran in "geany" (sudo apt-get install geany):
^[^ ]+ ?(.*$)

so i ended up with the following result:
INFO  [com.myapp.tiv.persistentservices.transportagreement.daoimpl....
INFO  [stdout] (http-/ setErrorMessage: Feil oppstod...
ERROR [stderr] (http-/ org.hibernate.LazyInitializat...
ERROR [stderr] (http-/     at org.hibernate.proxy.Abstr...

at first i thought it would be good enough with just this:
^[^ ]+ ?

but in geany (or is the regex just wrong?), the matching didn't stop at the end of the line so the output was like this:
 [stdout] (http-/ setErrorMessage: Feil oppstod...


^[^ ]+ ?(.*$)

first color (pink): start matching attempt at the beginning of the line
second color (purple): and match one or more characters that aren't a space (i.e. match until you meet a space)
third color (blue): and one additional single space
fourth color (aqua blue): then put everything else after that last additional space, to the end of the line, into a "capturing group" so we can reinsert it back into the line using the backreference, \1

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