Substitute:
x = 2/3tant
dx = (2dt)/(3cos^2t) dt
As then t = arctan ((3x)/2) we have that t in (-pi/2,pi/2)
we have:
int dx/(4+9x^2)^(1/2) = 2/3 int ( dt)/cos^2t 1/(4+9*4/9tan^2t)^(1/2) = 1/3 int ( dt)/cos^2t 1/(1+tan^2t)^(1/2)
Use now the trigonometric identity:
1+tan^2t = 1+sin^2t/cos^2t = (cos^2t+sin^2t)/cos^2t = 1/cos^2t
int dx/(4+9x^2)^(1/2) = 2/3int ( dt)/cos^2t 1/(1/cos^2t)^(1/2)
For t in (-pi/2,pi/2), cos t >0 so:
(1/cos^2t)^(1/2) = 1/cost
and:
int dx/(4+9x^2)^(1/2) = 2/3int ( dt)/cos^2t 1/(1/cost) = 2/3int (costdt)/cos^2t
we do not simplify but substitute again:
u=sint
du = cost dt
cos^2t = 1-sin^2t = 1-u^2
int dx/(4+9x^2)^(1/2) = 2/3 int (du)/(1-u^2)
which can be solved by partial fractions:
1/(1-u^2) = 1/((1+u)(1-u)) = 1/2 1/(1+u) + 1/2 1/(1-u)
so:
int (du)/(1-u^2) = 1/2int (du)/(1-u) +1/2 int (du)/(1+u) = -1/2lnabs(1-u)+1/2 ln abs (1+u)+C
and using the properties of logarithms:
int (du)/(1-u^2) = ln abs ((1+u)/(1-u))^(1/2) + C
Reversing the substitution we have:
u = sint = tant cost = tant/sqrt(1+tan^2t) = (3/2x)/sqrt(1+(3/2x)^2) = (3x)/sqrt(4+9x^2)
So:
int dx/(4+9x^2)^(1/2) = 2/3 ln sqrt ( (1 +((3x)/sqrt(4+9x^2)))/ ( 1 -((3x)/sqrt(4+9x^2)))) + C
int dx/(4+9x^2)^(1/2) = 2/3 ln sqrt ( (sqrt(4+9x^2)+3x)/( sqrt(4+9x^2) - 3x))+ C
Rationalize the denominator of the argument:
int dx/(4+9x^2)^(1/2) = 2/3 ln sqrt ( ((sqrt(4+9x^2)+3x)( sqrt(4+9x^2) + 3x))/ ( ( sqrt(4+9x^2) - 3x)( sqrt(4+9x^2) + 3x)))+ C
int dx/(4+9x^2)^(1/2) = 2/3 ln sqrt ( ((sqrt(4+9x^2)+3x)( sqrt(4+9x^2) + 3x))/ ( (4+9x^2-9x^2))+ C
int dx/(4+9x^2)^(1/2) = 2/3 ln sqrt ( (sqrt(4+9x^2)+3x)^2)/ sqrt( (4+9x^2-9x^2))+ C
int dx/(4+9x^2)^(1/2) = 2/3 ln sqrt ( (sqrt(4+9x^2)+3x)^2)/ sqrt( 4)+ C
int dx/(4+9x^2)^(1/2) = 1/3 ln abs ( (sqrt(4+9x^2)+3x))+ C
