What is int(lnx^9)/x^4 dx?

1 Answer
Nov 8, 2015

It is 9[-1/3x^-3 lnx -1/9 x^-3 +C] Which we may prefer to write:
-(3lnx+1)/x^3+C

Explanation:

lnx^9 = 9lnx, so

int lnx^9/x^2 dx = 9int x^(-4) lnx dx Now use integration by parts.

Let u=lnx and dv=x^-4 dx,

so du=x^(-1) dx and v = -1/3x^-3

And the integral becomes:

9[-1/3x^-3 lnx +1/3 int x^-3 x^-1 dx]

= 9[-1/3x^-3 lnx +1/3 int x^-4 dx]

= 9[-1/3x^-3 lnx -1/9 x^-3 +C]

Which we may prefer to write:

= -(3lnx+1)/x^3+C