Calculus: Early Transcendentals, Chapter 7, 7.3, Section 7.3, Problem 13

intsqrt(x^2-9)/x^3dx
Let's evaluate the integral: Apply integration by parts,
intuv'=uv-intu'v
Let u=sqrt(x^2-9)
v'=1/x^3
u'=d/dx(sqrt(x^2-9))
u'=1/2(x^2-9)^(1/2-1)(2x)
u'=x/sqrt(x^2-9)
v=int1/x^3dx
v=x^(-3+1)/(-3+1)
v=-1/(2x^2)
intsqrt(x^2-9)/x^3dx=sqrt(x^2-9)(-1/(2x^2))-intx/sqrt(x^2-9)(-1/(2x^2))dx
=-sqrt(x^2-9)/(2x^2)+1/2int1/(xsqrt(x^2-9))dx
Apply the integral substitution y=sqrt(x^2-9)
dy=1/2(x^2-9)^(1/2-1)(2x)dx
dy=x/sqrt(x^2-9)dx
dy=(xdx)/y
=-sqrt(x^2-9)/(2x^2)+1/2int(ydy)/x(1/(xy))
=-sqrt(x^2-9)/(2x^2)+1/2intdy/x^2
=-sqrt(x^2-9)/(2x^2)+1/2intdy/(y^2+9)
Now use the standard integral:int1/(x^2+a^2)dx=1/aarctan(x/a)+C
=-sqrt(x^2-9)/(2x^2)+1/2(1/3arctan(y/3))
Substitute back y=sqrt(x^2-9) and add a constant C to the solution,
=-sqrt(x^2-9)/(2x^2)+1/6arctan(sqrt(x^2-9)/3)+C