How to solve #int 1/(x+x^(2017)) dx# step by step?

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Answer 1 · 2018-03-19T16:22:04

#int dx/(x+x^2017) = -1/2016 ln (1/x^2016+1) +C#

#int dx/(x+x^2017) = int dx/(x^2017(1/x^2016+1)#

Substitute:

#u= (1/x^2016+1)#

#du = -2016/x^2017#

so:

#int dx/(x+x^2017) = -1/2016 int (du)/u#

#int dx/(x+x^2017) = -1/2016 ln abs u +C#

and undoing the substitution:

#int dx/(x+x^2017) = -1/2016 ln (1/x^2016+1) +C#