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x-2-1-dx-x-x-1-x-1-




Question Number 200403 by Anonim_X last updated on 18/Nov/23
             ∫  (((x^2  +  1)dx)/(x(x−1)(x+1))) = ??
$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\int\:\:\frac{\left(\boldsymbol{{x}}^{\mathrm{2}} \:+\:\:\mathrm{1}\right)\boldsymbol{{dx}}}{\boldsymbol{{x}}\left(\boldsymbol{{x}}−\mathrm{1}\right)\left(\boldsymbol{{x}}+\mathrm{1}\right)}\:=\:?? \\ $$$$ \\ $$
Answered by cortano12 last updated on 18/Nov/23
 I = ∫ ((x^2 −1+2)/(x(x−1)(x+1))) dx    I= ln ∣x∣ + 2 ∫ (dx/(x(x+1)(x−1)))    I= ln ∣x∣ + 2∫(− (1/x)+(1/(2(x+1)))+(1/(2(x−1))))dx   I = −ln ∣x∣+ln ∣x+1∣+ln ∣x−1∣ + c
$$\:\mathrm{I}\:=\:\int\:\frac{\mathrm{x}^{\mathrm{2}} −\mathrm{1}+\mathrm{2}}{\mathrm{x}\left(\mathrm{x}−\mathrm{1}\right)\left(\mathrm{x}+\mathrm{1}\right)}\:\mathrm{dx}\: \\ $$$$\:\mathrm{I}=\:\mathrm{ln}\:\mid\mathrm{x}\mid\:+\:\mathrm{2}\:\int\:\frac{\mathrm{dx}}{\mathrm{x}\left(\mathrm{x}+\mathrm{1}\right)\left(\mathrm{x}−\mathrm{1}\right)}\: \\ $$$$\:\mathrm{I}=\:\mathrm{ln}\:\mid\mathrm{x}\mid\:+\:\mathrm{2}\int\left(−\:\frac{\mathrm{1}}{\mathrm{x}}+\frac{\mathrm{1}}{\mathrm{2}\left(\mathrm{x}+\mathrm{1}\right)}+\frac{\mathrm{1}}{\mathrm{2}\left(\mathrm{x}−\mathrm{1}\right)}\right)\mathrm{dx} \\ $$$$\:\mathrm{I}\:=\:−\mathrm{ln}\:\mid\mathrm{x}\mid+\mathrm{ln}\:\mid\mathrm{x}+\mathrm{1}\mid+\mathrm{ln}\:\mid\mathrm{x}−\mathrm{1}\mid\:+\:\mathrm{c}\: \\ $$

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