`int sqrt(x^2 - 9)/x^3 dx` Evaluate the integral

`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`