Integral of dx /x^2018 - x=?

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Answer 1 · 2018-03-19T06:58:49

# 1/2017ln|(x^2017-1)/x^2017|+C, or ,#

# I=1/2017ln|(x^2017-1)|-ln|x|+C#.

Suppose that, #I=intdx/(x^2018-x)=intdx/{x(x^2017-1)#,

#=1/2017int(2017x^2016)/{x^2016*x(x^2017-1)dx#,

#=1/2017int(2017x^2016)/{x^2017(x^2017-1)dx#.

Now, we subst. #x^2017=y," so that, "2017x^2016dx=dy#.

#:. I=1/2017int1/{y(y-1)}dy#,

#=1/2017int{y-(y-1)}/{y(y-1)}dy#,

#=1/2017int{y/{y(y-1)}-(y-1)/{y(y-1)}}dy#,

#=1/2017int{1/(y-1)-1/y}dy#,

#=1/2017{ln|(y-1)|-ln|y|}#,

#=1/2017ln|(y-1)/y|#.

# rArr I=1/2017ln|(x^2017-1)/x^2017|+C, or ,#

# I=1/2017ln|(x^2017-1)|-ln|x|+C#.

#color(magenta)(BONUS)#

#int1/{x(x^n-1)}dx=1/nln|x^n-1|-ln|x|+c, (n!=0)#.