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

1 Answer
Nov 8, 2015

Answer:

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#