Question

How do you integrate `ln(x)/x^3`?

Answer 1

`int lnx /x^3 dx = (-lnx)/(2x^2 ) - 1/(4x^2) + C`

Explanation:

You should learn the IBP formula:
`int u(dv)/dxdx=uv - int v (du)/dxdx`

So essentially we are looking for one function that simplifies when it is differentiated, and one that simplifies when integrated (or at least is integrable).

Hopefully you can spot that `lnx` is not easy to integrate (you need to using IBP again), but is simpler when differentiated.

Let `{ (u=lnx, => , (du)/dx=1/x), ((dv)/dx=1/x^3=x^-3, =>, v=x^-2/-2 = -1/(2x^2) ) :}`

So IBP gives;

`int lnx 1/x^3 dx = (lnx)(-1/(2x^2) ) - int ( -1/(2x^2))(1/x)dx`
`:. int lnx /x^3 dx = (-lnx)/(2x^2 ) + 1/2 int ( 1/x^3 )dx`
`:. int lnx /x^3 dx = (-lnx)/(2x^2 ) + 1/2 (-1/(2x^2)) + C`
`:. int lnx /x^3 dx = (-lnx)/(2x^2 ) - 1/(4x^2) + C`