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

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Answer 1 · 2015-11-08T20:06:14

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

#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#