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#
