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Author: Tinku Tara

Question-211879

Question Number 211879 by Spillover last updated on 23/Sep/24 Answered by BHOOPENDRA last updated on 23/Sep/24 $$=\:\frac{\left({x}−\mathrm{4}\right)}{\mathrm{10}\left({x}^{\mathrm{2}} −\mathrm{4}{x}+\mathrm{5}\right)}\:+\frac{\mathrm{1}}{\mathrm{10}}\int\:\frac{{x}−\mathrm{4}}{{x}^{\mathrm{3}} −\mathrm{4}{x}^{\mathrm{2}} +\mathrm{5}{x}}\:\left({by}\:{ostrogradsky}'{s}\:{method}\right) \\ $$$$\:\:=\frac{{x}−\mathrm{4}}{\mathrm{10}\left({x}^{\mathrm{2}} −\mathrm{4}{x}+\mathrm{5}\right)}+\frac{\mathrm{1}}{\mathrm{10}}\left[\int\frac{\mathrm{4}{x}−\mathrm{11}}{\mathrm{5}\left({x}^{\mathrm{2}} −\mathrm{4}{x}+\mathrm{5}\right)}{dx}−\int\frac{\mathrm{4}}{\mathrm{5}{x}}\:{dx}\right] \\…

Question-211871

Question Number 211871 by Spillover last updated on 22/Sep/24 Answered by Ghisom last updated on 23/Sep/24 $$=−\underset{\mathrm{0}} {\overset{\mathrm{arccos}\:\frac{\sqrt{\mathrm{6}}}{\mathrm{3}}} {\int}}\frac{\mathrm{2}+\mathrm{tan}\:{x}}{\mathrm{3}−\mathrm{2cos}^{\mathrm{2}} \:{x}\:−\mathrm{sin}^{\mathrm{2}} \:{x}}{dx}= \\ $$$$\:\:\:\:\:\left[{t}=\sqrt{\mathrm{2}}\mathrm{tan}\:{x}\right] \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}}\underset{\mathrm{0}}…

ab-ba-4c-a-b-3c-

Question Number 211861 by hardmath last updated on 22/Sep/24 $$\overline {\mathrm{ab}}\:\:+\:\:\overline {\mathrm{ba}}\:\:=\:\:\mathrm{4c} \\ $$$$\mathrm{a}\:+\:\mathrm{b}\:+\:\mathrm{3c}\:=\:? \\ $$ Answered by A5T last updated on 22/Sep/24 $$\mathrm{10}{a}+{b}+\mathrm{10}{b}+{a}=\mathrm{11}\left({a}+{b}\right)=\mathrm{4}{c} \\…