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Question Number 85982 by Rio Michael last updated on 26/Mar/20

A primitive of the function defned by f (x) = x - 1 + \frac{1}{x + 1} is

A . F (x) = \frac{x^{2}}{2} - x + \ln (x + 1) B . F (x) = \frac{x^{2}}{2} + \ln (x - 1)

C . F (x) = \frac{x^{2}}{2} - x + \ln (1 - x) D . F (x) = - x + \ln (x - 1)

Commented by Serlea last updated on 26/Mar/20

Taking the integral, {it}^{'} s A

Commented by jagoll last updated on 26/Mar/20

F (x) = \int (x - 1 + \frac{1}{x + 1}) dx

= \frac{1}{2} x^{2} - x + \ln (x + 1)

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