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How do you integrate of #1/(1-x)#?

1 Answer

Jan 11, 2017

#int (dx)/(1-x) = -lnabs(1-x)#

Explanation:

Substitute:

#t= 1-x#
#dt = -dx#

and you have:

#int (dx)/(1-x) = -int (dt)/t = -ln abs (t) = -lnabs(1-x)#

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