Let J=∫(1/√(x2-9))dx.
1+tan2α=sec2α, so tan2α=sec2α-1, where α is any angle.
Use this identity: let x=3secθ, then x2=9sec2θ, and x2-9=9(sec2θ-1)=9tan2θ, √(x2-9)=3tanθ; dx/dθ=3secθtanθ.
J=∫(1/(3tanθ))(3secθtanθ)dθ=∫secθdθ.
We can write secθ as secθ(secθ+tanθ)/(secθ+tanθ)=(sec2θ+secθtanθ)/(secθ+tanθ). The numerator is the derivative of the denominator, so:
J=∫(sec2θ+secθtanθ)dθ/(secθ+tanθ)=ln|secθ+tanθ|+C.
Now can write this in terms of x, using secθ=⅓x, tanθ=⅓√(x2-9):
J=ln|⅓x+⅓√(x2-9)|+C. The constant C can be incorporated into the log argument:
J=ln|A(x+√(x2-9)|. A also absorbs the ⅓ factor.
CHECK
Differentiate ln|A(x+√(x2-9)|:
A(1+x/√(x2-9))/(A(x+√(x2-9))=(√(x2-9)+x)/[√(x2-9)(x+√(x2-9)]=1/√(x2-9), the original integrand.