What is #int(lnx^9)/x^4 dx#?
1 Answer
Nov 8, 2015
It is
Explanation:
Let
so
And the integral becomes:
#= 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#