inverse trig integration:two methods=two diff ans HELP!!

Q is whats the integral of (3)divided by(9x^2+6x+5) with no limits....

...method 1. ..............9( x^2 + (2/3)(x) + (5/9) )

=9( (x^2 + 1/3 )^2 + 4/9 )

therefore cancel 3 with 9 outside integral brackets to give 1/3 and times by

1/a i.e where a^2 = 4/9 which gives 1/2.....

.............1/2arctan 3/2(x^2 + 1/3 ) + c which gives the actual answer ....................1/2arctan(3x+1)/2 + c

method 2. which i see as perfectly ok please correct me........

(9x^2 +6x +5) simplifies to (3x + 1)^2 +4 ..........therefore taked 3 on top outside bracket and a^2 = 4.......and 3/2arctan(3x+1)/2...............

.........i dont understand why the mix up with 3/2 and 1/2 came from ....shouldn't happen..