Question Number 91990 by Power last updated on 04/May/20
Commented by mathmax by abdo last updated on 04/May/20
$${I}\:=\int\:\frac{{x}^{\mathrm{3}} −\mathrm{6}}{{x}^{\mathrm{4}} \:+\mathrm{6}{x}\:+\mathrm{8}}\:\Rightarrow\:{I}\:=\frac{\mathrm{1}}{\mathrm{4}}\int\:\:\frac{\mathrm{4}{x}^{\mathrm{3}} −\mathrm{24}}{{x}^{\mathrm{4}} \:+\mathrm{6}{x}\:+\mathrm{8}}{dx}\:\Rightarrow \\ $$$$\mathrm{4}{I}\:=\int\frac{\mathrm{4}{x}^{\mathrm{3}} \:+\mathrm{6}\:−\mathrm{30}}{{x}^{\mathrm{4}} \:+\mathrm{6}{x}\:+\mathrm{8}}{dx}\:={ln}\mid{x}^{\mathrm{4}} \:+\mathrm{6}{x}\:+\mathrm{8}\mid−\mathrm{30}\:\int\:\:\frac{{dx}}{{x}^{\mathrm{4}} \:+\mathrm{6}{x}\:+\mathrm{8}} \\ $$$${x}^{\mathrm{4}} \:+\mathrm{6}{x}\:+\mathrm{8}\:=\mathrm{0}\:{the}\:{roots}\:{are}\: \\ $$$${z}_{\mathrm{1}} =\mathrm{1},\mathrm{2402}\:+\mathrm{1},\mathrm{6576}{i}\:\:\left({complex}\right) \\ $$$${z}_{\mathrm{2}} =\mathrm{1},\mathrm{2402}\:−\mathrm{1},\mathrm{6576}{i}\:\left({complex}\right) \\ $$$${z}_{\mathrm{3}} =−\mathrm{1},\mathrm{2402}\:+\mathrm{0},\mathrm{5732}{i}\:\left({complex}\right) \\ $$$${z}_{\mathrm{4}} =−\mathrm{1},\mathrm{2402}\:−\mathrm{0},\mathrm{5732}{i}\:\left({complex}\right) \\ $$$${let}\:\alpha\:=\mathrm{1},\mathrm{2402}\:{and}\:\beta\:=\mathrm{0},\mathrm{5732}\:\Rightarrow{z}_{\mathrm{1}} =\alpha\:+\left(\mathrm{1}+\beta\right){i} \\ $$$${z}_{\mathrm{2}} =\alpha−\left(\mathrm{1}+\beta\right){i} \\ $$$${z}_{\mathrm{3}} =−\alpha\:+\beta{i}\:\:\:\:\:\:{and}\:{z}_{\mathrm{4}} =−\alpha−\beta{i}\:\Rightarrow \\ $$$$\int\:\:\frac{{dx}}{{x}^{\mathrm{4}} \:+\mathrm{6}{x}+\mathrm{8}}\:=\int\:\:\frac{{dx}}{\left({x}+\alpha−\beta{i}\right)\left({x}+\alpha+\beta{i}\right)\left({x}−\alpha+\left(\mathrm{1}+\beta\right){i}\right)\left({x}−\alpha−\left(\mathrm{1}+\beta\right){i}\right)} \\ $$$${but}\:\:\left({x}+\alpha−\beta{i}\right)\left({x}+\alpha+\beta{i}\right)\:=\left({x}−{z}_{\mathrm{3}} \right)\left({x}−\overset{−} {{z}}_{\mathrm{3}} \right)={x}^{\mathrm{2}} −\mathrm{2}{Re}\left({z}_{\mathrm{3}} \right){x}+\mid{z}_{\mathrm{3}} \mid^{\mathrm{2}} \\ $$$$={x}^{\mathrm{2}} +\mathrm{2}\alpha{x}\:+\alpha^{\mathrm{2}} \:+\beta^{\mathrm{2}} \\ $$$$\left({x}−{z}_{\mathrm{1}} \right)\left({x}−{z}_{\mathrm{2}} \right)\:=\left({x}−{z}_{\mathrm{1}} \right)\left({x}−\overset{−} {{z}}_{\mathrm{1}} \right)={x}^{\mathrm{2}} \:−\mathrm{2}{Re}\left({z}_{\mathrm{1}} \right){x}\:+\mid{z}_{\mathrm{1}} \mid^{\mathrm{2}} \\ $$$$={x}^{\mathrm{2}} −\mathrm{2}\alpha\:{x}\:+\alpha^{\mathrm{2}} \:+\left(\mathrm{1}+\beta\right)^{\mathrm{2}} \:\Rightarrow \\ $$$$\int\:\:\frac{{dx}}{{x}^{\mathrm{4}} \:+\mathrm{6}{x}\:+\mathrm{8}}\:\:=\int\:\:\:\frac{{dx}}{\left({x}^{\mathrm{2}} \:+\mathrm{2}\alpha{x}\:+\alpha^{\mathrm{2}} \:+\beta^{\mathrm{2}} \right)\left({x}^{\mathrm{2}} −\mathrm{2}\alpha{x}\:+\alpha^{\mathrm{2}} +\left(\mathrm{1}+\beta\right)^{\mathrm{2}} \right)} \\ $$$$…{be}\:{continued}…. \\ $$
Commented by Power last updated on 04/May/20
$$\mathrm{thanks} \\ $$