Question-91990

Question Number 91990 by Power last updated on 04/May/20

Commented by mathmax by abdo last updated on 04/May/20

I =∫ ((x^3 −6)/(x^4 +6x +8)) ⇒ I =(1/4)∫ ((4x^3 −24)/(x^4 +6x +8))dx ⇒ 4I =∫((4x^3 +6 −30)/(x^4 +6x +8))dx =ln∣x^4 +6x +8∣−30 ∫ (dx/(x^4 +6x +8)) x^4 +6x +8 =0 the roots are z_1 =1,2402 +1,6576i (complex) z_2 =1,2402 −1,6576i (complex) z_3 =−1,2402 +0,5732i (complex) z_4 =−1,2402 −0,5732i (complex) let α =1,2402 and β =0,5732 ⇒z_1 =α +(1+β)i z_2 =α−(1+β)i z_3 =−α +βi and z_4 =−α−βi ⇒ ∫ (dx/(x^4 +6x+8)) =∫ (dx/((x+α−βi)(x+α+βi)(x−α+(1+β)i)(x−α−(1+β)i))) but (x+α−βi)(x+α+βi) =(x−z_3 )(x−z_3 ^− )=x^2 −2Re(z_3 )x+∣z_3 ∣^2 =x^2 +2αx +α^2 +β^2 (x−z_1 )(x−z_2 ) =(x−z_1 )(x−z_1 ^− )=x^2 −2Re(z_1 )x +∣z_1 ∣^2 =x^2 −2α x +α^2 +(1+β)^2 ⇒ ∫ (dx/(x^4 +6x +8)) =∫ (dx/((x^2 +2αx +α^2 +β^2 )(x^2 −2αx +α^2 +(1+β)^2 ))) ...be continued....

$${I}\:=\int\:\frac{{x}^{\mathrm{3}} −\mathrm{6}}{{x}^{\mathrm{4}} \:+\mathrm{6}{x}\:+\mathrm{8}}\:\Rightarrow\:{I}\:=\frac{\mathrm{1}}{\mathrm{4}}\int\:\:\frac{\mathrm{4}{x}^{\mathrm{3}} −\mathrm{24}}{{x}^{\mathrm{4}} \:+\mathrm{6}{x}\:+\mathrm{8}}{dx}\:\Rightarrow \\ $$$$\mathrm{4}{I}\:=\int\frac{\mathrm{4}{x}^{\mathrm{3}} \:+\mathrm{6}\:−\mathrm{30}}{{x}^{\mathrm{4}} \:+\mathrm{6}{x}\:+\mathrm{8}}{dx}\:={ln}\mid{x}^{\mathrm{4}} \:+\mathrm{6}{x}\:+\mathrm{8}\mid−\mathrm{30}\:\int\:\:\frac{{dx}}{{x}^{\mathrm{4}} \:+\mathrm{6}{x}\:+\mathrm{8}} \\ $$$${x}^{\mathrm{4}} \:+\mathrm{6}{x}\:+\mathrm{8}\:=\mathrm{0}\:{the}\:{roots}\:{are}\: \\ $$$${z}_{\mathrm{1}} =\mathrm{1},\mathrm{2402}\:+\mathrm{1},\mathrm{6576}{i}\:\:\left({complex}\right) \\ $$$${z}_{\mathrm{2}} =\mathrm{1},\mathrm{2402}\:−\mathrm{1},\mathrm{6576}{i}\:\left({complex}\right) \\ $$$${z}_{\mathrm{3}} =−\mathrm{1},\mathrm{2402}\:+\mathrm{0},\mathrm{5732}{i}\:\left({complex}\right) \\ $$$${z}_{\mathrm{4}} =−\mathrm{1},\mathrm{2402}\:−\mathrm{0},\mathrm{5732}{i}\:\left({complex}\right) \\ $$$${let}\:\alpha\:=\mathrm{1},\mathrm{2402}\:{and}\:\beta\:=\mathrm{0},\mathrm{5732}\:\Rightarrow{z}_{\mathrm{1}} =\alpha\:+\left(\mathrm{1}+\beta\right){i} \\ $$$${z}_{\mathrm{2}} =\alpha−\left(\mathrm{1}+\beta\right){i} \\ $$$${z}_{\mathrm{3}} =−\alpha\:+\beta{i}\:\:\:\:\:\:{and}\:{z}_{\mathrm{4}} =−\alpha−\beta{i}\:\Rightarrow \\ $$$$\int\:\:\frac{{dx}}{{x}^{\mathrm{4}} \:+\mathrm{6}{x}+\mathrm{8}}\:=\int\:\:\frac{{dx}}{\left({x}+\alpha−\beta{i}\right)\left({x}+\alpha+\beta{i}\right)\left({x}−\alpha+\left(\mathrm{1}+\beta\right){i}\right)\left({x}−\alpha−\left(\mathrm{1}+\beta\right){i}\right)} \\ $$$${but}\:\:\left({x}+\alpha−\beta{i}\right)\left({x}+\alpha+\beta{i}\right)\:=\left({x}−{z}_{\mathrm{3}} \right)\left({x}−\overset{−} {{z}}_{\mathrm{3}} \right)={x}^{\mathrm{2}} −\mathrm{2}{Re}\left({z}_{\mathrm{3}} \right){x}+\mid{z}_{\mathrm{3}} \mid^{\mathrm{2}} \\ $$$$={x}^{\mathrm{2}} +\mathrm{2}\alpha{x}\:+\alpha^{\mathrm{2}} \:+\beta^{\mathrm{2}} \\ $$$$\left({x}−{z}_{\mathrm{1}} \right)\left({x}−{z}_{\mathrm{2}} \right)\:=\left({x}−{z}_{\mathrm{1}} \right)\left({x}−\overset{−} {{z}}_{\mathrm{1}} \right)={x}^{\mathrm{2}} \:−\mathrm{2}{Re}\left({z}_{\mathrm{1}} \right){x}\:+\mid{z}_{\mathrm{1}} \mid^{\mathrm{2}} \\ $$$$={x}^{\mathrm{2}} −\mathrm{2}\alpha\:{x}\:+\alpha^{\mathrm{2}} \:+\left(\mathrm{1}+\beta\right)^{\mathrm{2}} \:\Rightarrow \\ $$$$\int\:\:\frac{{dx}}{{x}^{\mathrm{4}} \:+\mathrm{6}{x}\:+\mathrm{8}}\:\:=\int\:\:\:\frac{{dx}}{\left({x}^{\mathrm{2}} \:+\mathrm{2}\alpha{x}\:+\alpha^{\mathrm{2}} \:+\beta^{\mathrm{2}} \right)\left({x}^{\mathrm{2}} −\mathrm{2}\alpha{x}\:+\alpha^{\mathrm{2}} +\left(\mathrm{1}+\beta\right)^{\mathrm{2}} \right)} \\ $$$$…{be}\:{continued}…. \\ $$

Commented by Power last updated on 04/May/20

$$\mathrm{thanks} \\ $$

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