How do you integrate square root of x^2-9 all if that over x^3 with respect to x?

Question

What is the answer?

Answer 1

Let x = 3secθ

 

Then dx = 3secθtanθdθ

 

So, ∫√(x²-9)/x³dx = ∫√(9sec²θ - 9)/ (3secθ)³ 3secθtanθdθ

 

                         = (1/3)∫√(sec²θ - 1) tanθ / sec²θdθ

 

                         = (1/3)∫√tan²θ tanθ/sec²θdθ

 

                         = (1/3)∫sin²θdθ

 

                         = (1/3)∫[(1-cos(2θ)/2]dθ

 

                         = 1/6[θ - (1/2)sin(2θ)] + C

 

                         = (1/6)θ - (1/12)(2sinθcosθ) + C

 

                         = (1/6)Arcsec(x/3) - (1/6)[√(x²-9)/x][3/x] + C

 

                         = (1/6)Arcsec (x/3) - (1/2)√(x²-9)/x² + C