evaluate the integral

Question

∫(1)/(√(x^2-9))dx

Answer 1

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.