As the discriminant of #9x^2+6x+5# is #6^2-4xx9xx5=36-180=-144#, it being negative, cannot be factorized in rational factors.
Observing that differential of #9x^2+6x+5# is #18x+6#, let us split #(x+1)/(9x^2+6x+5)# as #(18x+18)/(18(9x^2+6x+5))# or
#(18x+6)/(18(9x^2+6x+5))+12/(18(9x^2+6x+5))#
Hence #int(x+1)/(9x^2+6x+5)dx#
= #int(18x+6)/(18(9x^2+6x+5))dx+int12/(18(9x^2+6x+5))dx#
= #1/18int(18x+6)/(9x^2+6x+5)dx+2/3int1/(9x^2+6x+5)dx#
Let us integrate first part and assume #u=9x^2+6x+5#, then #du=(18x+6)dx# and it can be written as #1/18int(du)/u=1/18lnu=1/18ln(9x^2+6x+5)#.
For second part #2/3int1/(9x^2+6x+5)dx#, we will use the identity #int(dx)/(x^2+a^2)=1/atan^(-1)(x/a)+c#. For this converting denominator into sum of two squares, we get #(9x^2+6x+1)+2^2#, hence
#2/3int1/(9x^2+6x+5)dx=2/27int1/((x^2+2/3x+1/9)+(2/3)^2)dx=2/27int1/((x+1/3)^2+(2/3)^2)dx#
Let us now substitute #u=x+1/3# and as #du=dx# this is now
#2/27int1/(u^2+(2/3)^2)du=2/27xx(3/2tan^(-1)((3u)/2))# or
#1/9tan^(-1)((3(x+1/3))/2)=1/9tan^(-1)((3x+1)/2)#
Hence, #int(x+1)/(9x^2+6x+5)dx#
= #1/18ln(9x^2+6x+5)+1/9tan^(-1)((3x+1)/2)+c#
