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Question Number 32402 by saru53424@gmail.com last updated on 24/Mar/18

∫((x+2)/(1−x))

$$\int\frac{{x}+\mathrm{2}}{\mathrm{1}−{x}} \\ $$

Commented by abdo imad last updated on 24/Mar/18

∫((x+2)/(1−x))dx =−∫ ((x+2)/(x−1))dx =− ∫((x−1+3)/(x−1))dx  =−x −3∫ (dx/(x−1)) =−x −3ln∣x−1∣ +λ .

$$\int\frac{{x}+\mathrm{2}}{\mathrm{1}−{x}}{dx}\:=−\int\:\frac{{x}+\mathrm{2}}{{x}−\mathrm{1}}{dx}\:=−\:\int\frac{{x}−\mathrm{1}+\mathrm{3}}{{x}−\mathrm{1}}{dx} \\ $$$$=−{x}\:−\mathrm{3}\int\:\frac{{dx}}{{x}−\mathrm{1}}\:=−{x}\:−\mathrm{3}{ln}\mid{x}−\mathrm{1}\mid\:+\lambda\:. \\ $$

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