#int dx/(xsqrt(16x^2-9)) = int dx/(3xsqrt ((4/3x)^2 -1))#
Substitute:
#4/3 x = sect#, #dx = 3/4 sect tant dt#
and restrict for now to #x in (3/4,+oo)# so that #t in (0,pi/2)#
#int dx/(xsqrt(16x^2-9)) = 3/4 int (sect tant dt)/(9/4 sect sqrt(sec^2t-1))#
#int dx/(xsqrt(16x^2-9)) = 1/3 int ( tant dt)/ sqrt(sec^2t-1)#
Use now the trigonometric identity:
#sec^2t -1 = 1/cos^2t -1 = (1-sin^2t)/cos^2t = tan^2t#
As #t in (0,pi/2) => tant >=0#
then:
#sqrt(sec^2t-1) = tant#
and:
#int dx/(xsqrt(16x^2-9)) = 1/3 int ( tant dt)/ tant = 1/3 int dt =t/3 +C#
and undoing the substitution:
#int dx/(xsqrt(16x^2-9)) = 1/3 arcsec(4/3x)+C#
For #x in(-oo,-3/4)# we can use the same procedure except that:
#sqrt(sec^2t-1) = -tant#
and accordingly:
#int dx/(xsqrt(16x^2-9)) = -1/3 arcsec(4/3x)+C#
So we can express the integral for any #x# as:
#int dx/(xsqrt(16x^2-9)) = x/(3absx) arcsec(4/3x)+C#
Note that as:
#arcsec x = arctan (xsqrt((x^2-1)/x^2))#
We can also express the integral as:
#int dx/(xsqrt(16x^2-9)) = x/(3absx) arctan ((4x)/3 sqrt (( 16x^2-9)/(16x^2))) +C#
#int dx/(xsqrt(16x^2-9)) = x/(3absx) arctan ( sqrt ( 16x^2-9)/3) +C#
