What is #int (1+x)/(1-x) dx# ?

1 Answer
George C.
Jan 9, 2018

#int (1+x)/(1-x) dx = -2ln abs(x-1)-x+C#

Explanation:

#int (1+x)/(1-x) dx = int (2-(1-x))/(1-x))dx#

#color(white)(int (1+x)/(1-x) dx) = int (2/(1-x)-1) dx#

#color(white)(int (1+x)/(1-x) dx) = int (-2/(x-1)-1) dx#

#color(white)(int (1+x)/(1-x) dx) = -2ln abs(x-1)-x+C#

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