1/2[6ln(x+1)+ln(x-2)]?
What about it ?
We could change it to ....
1/2[ln(x+1)^6+ln(x-2)]
= 1/2[ln ( (x+1)^6/(x-2) )
= ln √((x+1)^6/(x-2))
but why ?
We could change it to ....
1/2[ln(x+1)^6+ln(x-2)]
= 1/2[ln ( (x+1)^6/(x-2) )
= ln √((x+1)^6/(x-2))
but why ?