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

$$\mathrm{A}\mathrm{primitive}\mathrm{of}\mathrm{the}\mathrm{function}\mathrm{defned}\mathrm{by}\mathrm{f}\left(x\right)=x-1+\frac{1}{x+1}\mathrm{is}$$$$\mathrm{A}.\mathrm{F}\left(x\right)=\frac{x^2}{2}-x+\ln (x+1)\mathrm{B}.\mathrm{F}\left(x\right)=\frac{x^2}{2}+\ln (x-1)$$$$\mathrm{C}.\mathrm{F}\left(x\right)=\frac{x^2}{2}-x+\ln (1-x)\mathrm{D}.\mathrm{F}\left(x\right)=-x+\ln (x-1)$$$$$$

Commented by Serlea last updated on 26/Mar/20

$$\mathrm{Taking}\mathrm{the}\mathrm{integral},{\mathrm{it}}^{\prime }\mathrm{s}\mathrm{A}$$

Commented by jagoll last updated on 26/Mar/20

$$\mathrm{F}\left(\mathrm{x}\right)=\int (\mathrm{x}-1+\frac{1}{\mathrm{x}+1})\mathrm{dx}$$$$=\frac{1}{2}{\mathrm{x}}^2-\mathrm{x}+\ln (\mathrm{x}+1)$$

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