Instead of usual approach with Weierstrass substitution, we can simplify the denominator as the following;
#2sinx+3cosx=Acos(x+phi)#
But we have to find #A# and #phi#... Well, we can expand the #cos(x+phi)# to get:
#color(red)2sinx+color(red)3cosx=color(red)(-Asinphi)sinx+color(red)(Acosphi)cosx#
In order for our relation to be true, we must have
#Asinphi=-2#
#Acosphi = 3#
To find #A#, square both relations and add them.
#=> A^2 sin^2phi + A^2cos^2phi = 13#
Taking #A^2# as common factor, the #sin^2phi + cos^2phi# simplifies to #1#.
#=> A^2 = 13 => A=sqrt13#, taking only the positive root.
We don't need to find #phi# for now, since it's just a constant.
#:. int 1/(2sinx+3cosx)dx = 1/sqrt(13) int sec(x+phi) dx#
We'll solve this new integral by substitution.
Let #u = x+phi#, with #du=dx#.
#int sec(x+phi)dx = int secu du#
This is a standard integral:
#int secudu = ln|tanu+secu|+C#
By undoing the substitution, we get:
#1/sqrt(13) int sec(x+phi)dx = 1/sqrt13[ln|tan(x+phi)+sec(x+phi)|+C]#
Let #c = C/sqrt13#.
#1/sqrt13 intsec(x+phi)dx = 1/sqrt13ln|tan(x+phi)+sec(x+phi)|+c#
#:.int 1/(2sinx+3cosx)dx =1/sqrt13ln|tan(x+phi)+sec(x+phi)|+c#
Only thing left is to find #phi#. From our previous relations, we have:
#sinphi = -2/sqrt13#
#cosphi=3/sqrt13#
#=> tanphi = -2/3#
#color(red)(phi = arctan (-2/3)#
Thus,
#color(Red)(int 1/(2sinx+3cosx)dx#
#color(red)(=1/sqrt13ln|tan(x+phi)+sec(x+phi)|+c#
where #phi = arctan (-2/3)#.
