Calculus of a Single Variable, Chapter 8, 8.1, Section 8.1, Problem 56

y'=1/(xsqrt(4x^2-9))
y=int1/(xsqrt(4x^2-9))dx
Apply integral substitution: x=3/2sec(u)
dx=3/2sec(u)tan(u)du
y=int1/(3/2sec(u)sqrt(4(3/2sec(u))^2-9))(3/2sec(u)tan(u))du
y=inttan(u)/sqrt(4(9/4sec^2(u))-9)du
y=inttan(u)/sqrt(9sec^2(u)-9)du
y=inttan(u)/(sqrt(9)sqrt(sec^2(u)-1))du
Now use the identity:sec^2(x)=1+tan^2(x)
y=inttan(u)/(3sqrt(1+tan^2(u)-1))du
y=inttan(u)/(3sqrt(tan^2(u)))du
y=inttan(u)/(3tan(u))du assuming tan(u) >=0
y=int1/3du
take the constant out,
y=1/3intdu
y=1/3u
Substitute back u=arcsec((2x)/3)
and add a constant C to the solution,
y=1/3arcsec((2x)/3)+C