Complete the square at the denominator:
#9x^2-6x+5 = 9x^2-6x+1+4 = (3x-1)^2+4#
then write the integrand as:
#1/sqrt(9x^2-6x+5 ) = 1/sqrt( (3x-1)^2+4) = 1/2 1/sqrt(((3x-1)/2)^2+1)#
Substitute now:
#(3x-1)/2 = tant#
with #t in (-pi/2,pi/2)#
#dx = 2/3sec^2t dt#
so that:
#int dx/sqrt(9x^2-6x+5 ) = 1/3 int (sec^2tdt)/sqrt(tan^2t+1)#
Use now the trigonometric identity:
#tan^2t+1 = sec^2t#
and as for #t in (-pi/2,pi/2)# the secant is positive:
#sqrt(tan^2t+1) = sec t#
So:
#int dx/sqrt(9x^2-6x+5 ) = 1/3 int (sec^2tdt)/sect#
#int dx/sqrt(9x^2-6x+5 ) = 1/3 int sectdt#
#int dx/sqrt(9x^2-6x+5 ) = 1/3 ln abs(sect + tant) +C#
Undo the substitution, considering that:
#sect = sqrt(tan^2t +1) = sqrt(((3x-1)/2)^2+1) = 1/2sqrt(9x^2-6x+5 )#
#int dx/sqrt(9x^2-6x+5 ) = 1/3 ln abs(1/2sqrt(9x^2-6x+5 )+(3x-1)/2) +C#
and as multiplying the argument of the logarithm by a constant is equivalent to adding a constant to the expression:
#int dx/sqrt(9x^2-6x+5 ) = 1/3 ln abs(3x-1+sqrt(9x^2-6x+5 )) +C#
