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Question Number 116815 by bemath last updated on 07/Oct/20
∫ (dx/((x−2)(x^2 +4))) =?
$$\int\:\frac{\mathrm{dx}}{\left(\mathrm{x}−\mathrm{2}\right)\left(\mathrm{x}^{\mathrm{2}} +\mathrm{4}\right)}\:=? \\ $$
Answered by john santu last updated on 07/Oct/20
⇒ (1/((x−2)(x^2 +4))) = (A/(x−2)) + ((Bx+C)/(x^2 +4)) ⇒1 = A(x^2 +4)+(x−2)(Bx+C) put x=2⇒1 = 8A; A = (1/8) put x=0⇒1=4.(1/8)−2C ; 2C=−(1/2) C=−(1/4) put x=1⇒1=(5/8) −(B−(1/2)) B = (5/8) −(1/8) = (1/8) The integral becomes I = ∫ (1/(8x)) dx +∫(((1/8)x−(1/4))/(x^2 +4)) dx I= (1/8) ln ∣x∣ +(1/8)∫ ((x−2)/(x^2 +4)) dx I=(1/8)ln ∣x∣+(1/(16))∫ ((d(x^2 +4))/(x^2 +4))−(1/4)∫ (dx/(x^2 +4)) I=(1/8) ln ∣x∣ +(1/(16))ln ∣x^2 +4∣−(1/8) tan^(−1) ((x/2)) + c
$$\Rightarrow\:\frac{\mathrm{1}}{\left({x}−\mathrm{2}\right)\left({x}^{\mathrm{2}} +\mathrm{4}\right)}\:=\:\frac{{A}}{{x}−\mathrm{2}}\:+\:\frac{{Bx}+{C}}{{x}^{\mathrm{2}} +\mathrm{4}} \\ $$$$\Rightarrow\mathrm{1}\:=\:{A}\left({x}^{\mathrm{2}} +\mathrm{4}\right)+\left({x}−\mathrm{2}\right)\left({Bx}+{C}\right) \\ $$$${put}\:{x}=\mathrm{2}\Rightarrow\mathrm{1}\:=\:\mathrm{8}{A};\:{A}\:=\:\frac{\mathrm{1}}{\mathrm{8}} \\ $$$${put}\:{x}=\mathrm{0}\Rightarrow\mathrm{1}=\mathrm{4}.\frac{\mathrm{1}}{\mathrm{8}}−\mathrm{2}{C}\:;\:\mathrm{2}{C}=−\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{C}=−\frac{\mathrm{1}}{\mathrm{4}} \\ $$$${put}\:{x}=\mathrm{1}\Rightarrow\mathrm{1}=\frac{\mathrm{5}}{\mathrm{8}}\:−\left({B}−\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{B}\:=\:\frac{\mathrm{5}}{\mathrm{8}}\:−\frac{\mathrm{1}}{\mathrm{8}}\:=\:\frac{\mathrm{1}}{\mathrm{8}} \\ $$$${The}\:{integral}\:{becomes}\: \\ $$$${I}\:=\:\int\:\frac{\mathrm{1}}{\mathrm{8}{x}}\:{dx}\:+\int\frac{\frac{\mathrm{1}}{\mathrm{8}}{x}−\frac{\mathrm{1}}{\mathrm{4}}}{{x}^{\mathrm{2}} +\mathrm{4}}\:{dx} \\ $$$${I}=\:\frac{\mathrm{1}}{\mathrm{8}}\:\mathrm{ln}\:\mid{x}\mid\:+\frac{\mathrm{1}}{\mathrm{8}}\int\:\frac{{x}−\mathrm{2}}{{x}^{\mathrm{2}} +\mathrm{4}}\:{dx}\: \\ $$$${I}=\frac{\mathrm{1}}{\mathrm{8}}\mathrm{ln}\:\mid{x}\mid+\frac{\mathrm{1}}{\mathrm{16}}\int\:\frac{{d}\left({x}^{\mathrm{2}} +\mathrm{4}\right)}{{x}^{\mathrm{2}} +\mathrm{4}}−\frac{\mathrm{1}}{\mathrm{4}}\int\:\frac{{dx}}{{x}^{\mathrm{2}} +\mathrm{4}} \\ $$$${I}=\frac{\mathrm{1}}{\mathrm{8}}\:\mathrm{ln}\:\mid{x}\mid\:+\frac{\mathrm{1}}{\mathrm{16}}\mathrm{ln}\:\mid{x}^{\mathrm{2}} +\mathrm{4}\mid−\frac{\mathrm{1}}{\mathrm{8}}\:\mathrm{tan}^{−\mathrm{1}} \left(\frac{{x}}{\mathrm{2}}\right)\:+\:{c} \\ $$
Answered by Dwaipayan Shikari last updated on 07/Oct/20
∫(dx/((x−2)(x^2 +4))) =(1/8)∫(1/(x−2))−((x+2)/(x^2 +4))dx =(1/8)log(x−2)−(1/8)∫(x/(x^2 +4))−(1/4)∫(1/(x^2 +4))dx =(1/8)log(x−2)−(1/(16))log(x^2 +4)−(1/8)tan^(−1) (x/2)+C
$$\int\frac{{dx}}{\left({x}−\mathrm{2}\right)\left({x}^{\mathrm{2}} +\mathrm{4}\right)} \\ $$$$=\frac{\mathrm{1}}{\mathrm{8}}\int\frac{\mathrm{1}}{{x}−\mathrm{2}}−\frac{{x}+\mathrm{2}}{{x}^{\mathrm{2}} +\mathrm{4}}{dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{8}}{log}\left({x}−\mathrm{2}\right)−\frac{\mathrm{1}}{\mathrm{8}}\int\frac{{x}}{{x}^{\mathrm{2}} +\mathrm{4}}−\frac{\mathrm{1}}{\mathrm{4}}\int\frac{\mathrm{1}}{{x}^{\mathrm{2}} +\mathrm{4}}{dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{8}}{log}\left({x}−\mathrm{2}\right)−\frac{\mathrm{1}}{\mathrm{16}}{log}\left({x}^{\mathrm{2}} +\mathrm{4}\right)−\frac{\mathrm{1}}{\mathrm{8}}{tan}^{−\mathrm{1}} \frac{{x}}{\mathrm{2}}+{C} \\ $$

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