= 1/2 ln|( 1+2lnx)|+c
Original formula =1/3 ∫1√ (2+3 lnx) d (2+3 lnx)
=2/3 √(2+3lnx)+c
Original formula =-∫ 1/lncosx dlncosx
=-ln|lncosx|+c
Original formula =∫x/( 1+x? )dx-∫√arctanx/( 1+x? )dx
= 1/2∫ 1/( 1+x? )d( 1+x? )-∫√ Arctangent
= 1/2ln( 1+x? )-2/3 (arctanx)^(3/2)+c
Original formula =∫ 1/(4+ln? x) dlnx
= 1/2 south pole [(lnx)/2] +c