Question Number 221588 by Nicholas666 last updated on 08/Jun/25
$$ \\ $$$$\:\:\:\:\int\:\frac{\mathrm{8}{t}\:−\:\mathrm{8}{t}^{\:\mathrm{3}} }{{t}^{\:\mathrm{6}} \:+\:\mathrm{6}{t}^{\mathrm{5}} \:+\:\mathrm{3}{t}^{\:\mathrm{4}} \:−\:\mathrm{20}{t}^{\mathrm{3}} \:+\:\mathrm{3}{t}^{\mathrm{2}} \:+\:\mathrm{6}{t}\:+\:\mathrm{1}}\:{dt}\:\:\:\: \\ $$$$ \\ $$
Answered by Frix last updated on 08/Jun/25
$$=−\mathrm{8}\int\frac{{t}\left({t}−\mathrm{1}\right)\left({t}+\mathrm{1}\right)}{\left({t}−\mathrm{1}\right)^{\mathrm{2}} \left({t}^{\mathrm{2}} +\mathrm{4}{t}+\mathrm{1}\right)^{\mathrm{2}} }{dt}= \\ $$$$=−\mathrm{8}\int\frac{{t}\left({t}+\mathrm{1}\right)}{\left({t}−\mathrm{1}\right)\left({t}^{\mathrm{2}} +\mathrm{4}{t}+\mathrm{1}\right)^{\mathrm{2}} }{dt} \\ $$$$\mathrm{Now}\:\mathrm{decompose}\:\mathrm{etc}. \\ $$$$\mathrm{I}\:\mathrm{get} \\ $$$$\frac{\mathrm{2}}{\mathrm{9}}\mathrm{ln}\:\frac{{t}^{\mathrm{2}} +\mathrm{4}{t}+\mathrm{1}}{\left({t}−\mathrm{1}\right)^{\mathrm{2}} }\:−\frac{\mathrm{4}{t}}{\mathrm{3}\left({t}^{\mathrm{2}} +\mathrm{4}{t}+\mathrm{1}\right)}+{C} \\ $$