# What is the integral `int ln x/x^4 dx` ?

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### 1 Answer

The integral `int ln x/x^4 dx` has to be determined. Use integration by parts which gives `int u*dv = u*v - int v du`

Let `u = ln x => du = (1/x)*dx`

and `dv = (1/x^4)dx => v = -x^-3/3 = -1/(3*x^3)`

`int ln x/x^4 dx`

=> `-ln x/(3*x^3) - int -1/(3*x^3)*(1/x)*dx`

=> `-ln x/(3*x^3) - (1/3)*int -1/x^4*dx`

=> `-ln x/(3*x^3) - (1/3)*(-1)*(x^-3)/-3`

=> `-ln x/(3*x^3) - 1/(9*x^3) + C`

=> `(-3*ln x + 1)/(9*x^3) + C`

**The integral** `int ln x/x^4 dx = (-3*ln x + 1)/(9*x^3) + C`