Question Number 85362 by sahnaz last updated on 21/Mar/20 $$\int\frac{−\mathrm{4}−\mathrm{u}}{\mathrm{u}^{\mathrm{2}} +\mathrm{5u}+\mathrm{6}}\mathrm{du} \\ $$ Answered by john santu last updated on 22/Mar/20 $$\int\:\frac{{du}}{{u}+\mathrm{3}}\:−\int\:\frac{\mathrm{2}}{{u}+\mathrm{2}}\:{du}\:=\: \\ $$$$\:\mathrm{ln}\:\mid{u}+\mathrm{3}\mid\:−\:\mathrm{2}\:\mathrm{ln}\:\mid{u}+\mathrm{2}\mid\:+\:{c}\: \\…
Question Number 85360 by sahnaz last updated on 21/Mar/20 $$\int\left(\mathrm{e}^{\left(\mathrm{1}−\mathrm{x}\right)×\mathrm{e}^{\mathrm{x}} } ×\mathrm{e}^{\int\mathrm{xe}^{\mathrm{x}} \mathrm{dx}} \right)\mathrm{dx} \\ $$ Answered by john santu last updated on 21/Mar/20 $$\int\:{xe}^{{x}}…
Question Number 85355 by TawaTawa1 last updated on 21/Mar/20 Commented by mathmax by abdo last updated on 21/Mar/20 $${I}\:=\int\:{arctan}\left(\sqrt{\mathrm{1}+\sqrt{{x}}}\right){dx}\:\:{changement}\:\sqrt{\mathrm{1}+\sqrt{{x}}}={t}\:{give} \\ $$$$\mathrm{1}+\sqrt{{x}}={t}^{\mathrm{2}} \:\Rightarrow\sqrt{{x}}={t}^{\mathrm{2}} −\mathrm{1}\:\Rightarrow{x}\:=\left({t}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} \:\Rightarrow{dx}\:=\mathrm{2}\left(\mathrm{2}{t}\right)\left({t}^{\mathrm{2}}…
Question Number 150871 by liberty last updated on 16/Aug/21 $$\:\mathrm{Find}\:\mathrm{all}\:\mathrm{the}\:\mathrm{real}\:\mathrm{solutions} \\ $$$$\mathrm{of}\:\mathrm{cos}\:\mathrm{x}+\mathrm{cos}\:^{\mathrm{5}} \mathrm{x}+\mathrm{cos}\:\mathrm{7x}=\mathrm{3} \\ $$ Answered by EDWIN88 last updated on 16/Aug/21 $$\:\mathrm{cos}\:\mathrm{7}{x}+\mathrm{cos}\:{x}=\mathrm{2cos}\:\mathrm{4}{x}\:\mathrm{cos}\:\mathrm{3}{x} \\ $$$$\:=\:\mathrm{2}\left(\mathrm{2cos}\:^{\mathrm{2}}…
Question Number 85323 by sahnaz last updated on 20/Mar/20 $$\int\frac{\mathrm{z}−\mathrm{3}}{\mathrm{5z}−\mathrm{10}}\mathrm{dz} \\ $$ Answered by MJS last updated on 20/Mar/20 $$\int\frac{{z}−\mathrm{3}}{\mathrm{5}{z}−\mathrm{10}}{dz}=\frac{\mathrm{1}}{\mathrm{5}}\int\frac{{z}−\mathrm{3}}{{z}−\mathrm{2}}{dz}= \\ $$$$=\frac{\mathrm{1}}{\mathrm{5}}\int{dz}−\frac{\mathrm{1}}{\mathrm{5}}\int\frac{{dz}}{{z}−\mathrm{2}}=\frac{\mathrm{1}}{\mathrm{5}}{z}−\frac{\mathrm{1}}{\mathrm{5}}\mathrm{ln}\:\mid{z}−\mathrm{2}\mid\:+{C} \\ $$ Terms…
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Question Number 85319 by sahnaz last updated on 20/Mar/20 $$\int\frac{\mathrm{2}+\mathrm{u}}{−\mathrm{u}^{\mathrm{2}} −\mathrm{u}}\mathrm{du} \\ $$ Answered by MJS last updated on 20/Mar/20 $$\int\frac{\mathrm{2}+{u}}{−{u}^{\mathrm{2}} −{u}}{du}=−\int\frac{{u}+\mathrm{2}}{{u}\left({u}+\mathrm{1}\right)}{du}= \\ $$$$=\int\frac{{du}}{{u}+\mathrm{1}}−\mathrm{2}\int\frac{{du}}{{u}}=\mathrm{ln}\:\mid{u}+\mathrm{1}\mid\:−\mathrm{2ln}\:\mid{u}\mid\:+{C} \\…
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Question Number 85260 by Umar last updated on 20/Mar/20 $$\mathrm{Evaluate}\:\mathrm{using}\:\mathrm{cauchy}'\mathrm{s}\:\mathrm{integral}\: \\ $$$$\:\:\:\:\:\:\:\:\:\int_{\mathrm{c}} \:\frac{\mathrm{e}^{\mathrm{i}\pi} }{\left(\mathrm{z}^{\mathrm{2}} +\mathrm{4}\right)^{\mathrm{2}} \left(\mathrm{z}+\mathrm{1}\right)^{\mathrm{2}} }\mathrm{dz} \\ $$$$\mathrm{where}\:\mathrm{c}\:\mathrm{is}\:\mathrm{a}\:\mathrm{circle}\:\mathrm{with}\:\mid\mathrm{z}−\mathrm{i}\mid=\mathrm{3}.\mathrm{5} \\ $$$$ \\ $$$$\mathrm{help}\:\mathrm{please} \\ $$…
Question Number 85256 by john santu last updated on 20/Mar/20 $$\int\underset{\mathrm{0}} {\overset{\:\mathrm{1}} {\:}}\:\frac{\mathrm{sin}\:\left(\mathrm{ln}\:\mathrm{x}\right)}{\mathrm{ln}\:\left(\mathrm{x}\right)}\:\mathrm{dx}\: \\ $$ Commented by john santu last updated on 20/Mar/20 $$\mathrm{sin}\:\left(\mathrm{z}\right)\:=\:\frac{\mathrm{e}^{\mathrm{iz}\:} −\mathrm{e}^{−\mathrm{iz}}…