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

Question

Emma Aldridge · Accepted Answer

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

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#

Luke Alvarez · Answer

undefined To integrate ln(x)/x^3, you can use integration by parts. Let u = ln(x) and dv = 1/x^3 dx. Then differentiate u to get du = (1/x) dx, and integrate dv to get v = -1/(2x^2).

Now apply the integration by parts formula:

∫u dv = uv - ∫v du

Substituting the values we found:

∫ln(x)/x^3 dx = -ln(x)/(2x^2) - ∫(-1/(2x^2))(1/x) dx

Simplify the integral:

∫ln(x)/x^3 dx = -ln(x)/(2x^2) + 1/(2x^3) ∫ dx

Integrate 1/(2x^3) to get (-1/(4x^2)), and combine terms:

∫ln(x)/x^3 dx = -ln(x)/(2x^2) - 1/(4x^2) + C

Where C is the constant of integration.