= 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