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




Question Number 25865 by anonymous_ last updated on 15/Dec/17
∫((2+2x)/((x−1)(x^2 +1)))dx
$$\int\frac{\mathrm{2}+\mathrm{2}{x}}{\left({x}−\mathrm{1}\right)\left({x}^{\mathrm{2}} +\mathrm{1}\right)}{dx} \\ $$
Answered by ajfour last updated on 16/Dec/17
((2+2x)/((x−1)(x^2 +1)))=(A/(x−1))+((Bx+C)/(x^2 +1))  ⇒2+2x=A(x^2 +1)+(x−1)(Bx+C)  ⇒ A+B=0, C−B=2 , A−C=2  from first two  C+A=2  so  A=2, C=0, B=−2  ∫((2+2x)/((x−1)(x^2 −1)))dx =      2∫(dx/(x−1))−∫((2xdx)/(x^2 +1))      =2ln ∣x−1∣−ln ∣x^2 +1∣+c .
$$\frac{\mathrm{2}+\mathrm{2}{x}}{\left({x}−\mathrm{1}\right)\left({x}^{\mathrm{2}} +\mathrm{1}\right)}=\frac{{A}}{{x}−\mathrm{1}}+\frac{{Bx}+{C}}{{x}^{\mathrm{2}} +\mathrm{1}} \\ $$$$\Rightarrow\mathrm{2}+\mathrm{2}{x}={A}\left({x}^{\mathrm{2}} +\mathrm{1}\right)+\left({x}−\mathrm{1}\right)\left({Bx}+{C}\right) \\ $$$$\Rightarrow\:{A}+{B}=\mathrm{0},\:{C}−{B}=\mathrm{2}\:,\:{A}−{C}=\mathrm{2} \\ $$$${from}\:{first}\:{two}\:\:{C}+{A}=\mathrm{2} \\ $$$${so}\:\:{A}=\mathrm{2},\:{C}=\mathrm{0},\:{B}=−\mathrm{2} \\ $$$$\int\frac{\mathrm{2}+\mathrm{2}{x}}{\left({x}−\mathrm{1}\right)\left({x}^{\mathrm{2}} −\mathrm{1}\right)}{dx}\:= \\ $$$$\:\:\:\:\mathrm{2}\int\frac{{dx}}{{x}−\mathrm{1}}−\int\frac{\mathrm{2}{xdx}}{{x}^{\mathrm{2}} +\mathrm{1}} \\ $$$$\:\:\:\:=\mathrm{2ln}\:\mid{x}−\mathrm{1}\mid−\mathrm{ln}\:\mid{x}^{\mathrm{2}} +\mathrm{1}\mid+{c}\:. \\ $$

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