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Question-131334

Question Number 131334 by Algoritm last updated on 03/Feb/21 Answered by bluberry508 last updated on 06/Feb/21 $$ \\ $$$${let}\:\:{g}\left({x}\right)=\int_{\mathrm{0}} ^{\:{x}} {xf}\:'\left({t}\right)−{tf}^{\:'} \left({t}\right){dt}\:= \\ $$$${x}\int_{\mathrm{0}} ^{\:{x}}…

Question-131320

Question Number 131320 by mohammad17 last updated on 03/Feb/21 Answered by mr W last updated on 04/Feb/21 $$\mathrm{2}:\:\mathrm{1}+\mathrm{1}\:\Rightarrow\frac{\mathrm{1}}{\mathrm{36}} \\ $$$$\mathrm{3}:\:\mathrm{1}+\mathrm{2},\:\mathrm{2}+\mathrm{1}\:\Rightarrow\frac{\mathrm{2}}{\mathrm{36}}=\frac{\mathrm{1}}{\mathrm{18}} \\ $$$$\mathrm{4}:\:\mathrm{2}+\mathrm{2},\:\mathrm{3}+\mathrm{1},\:\mathrm{1}+\mathrm{3}\:\Rightarrow\frac{\mathrm{3}}{\mathrm{36}}=\frac{\mathrm{1}}{\mathrm{12}} \\ $$$$\mathrm{5}:\:\mathrm{2}+\mathrm{3},\:\mathrm{3}+\mathrm{2},\:\mathrm{1}+\mathrm{4},\:\mathrm{4}+\mathrm{1}\:\Rightarrow\frac{\mathrm{4}}{\mathrm{36}}=\frac{\mathrm{1}}{\mathrm{9}} \\…

Question-131305

Question Number 131305 by Algoritm last updated on 03/Feb/21 Answered by EDWIN88 last updated on 03/Feb/21 $$\left(\mathrm{8}\right)\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{cos}\:{x}−\mathrm{cos}\:\left(\mathrm{3}{x}\right)}{\mathrm{ln}\:\left(\mathrm{1}+{x}^{\mathrm{2}} \right)}\:=\: \\ $$$$\:\:\:\:\:\:\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\left(\mathrm{1}−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\right)−\left(\mathrm{1}−\frac{\mathrm{9}{x}^{\mathrm{2}} }{\mathrm{2}}\right)}{{x}^{\mathrm{2}} }\:=\:\mathrm{4}\:…

0-pi-2-1-9-cos-x-12-sin-x-dx-

Question Number 65760 by mmkkmm000m last updated on 03/Aug/19 $$\underset{\:\mathrm{0}} {\overset{\pi/\mathrm{2}} {\int}}\:\frac{\mathrm{1}}{\mathrm{9}\:\mathrm{cos}\:{x}+\mathrm{12}\:\mathrm{sin}\:{x}}\:{dx}\:= \\ $$ Commented by mathmax by abdo last updated on 04/Aug/19 $${let}\:{I}\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}}…

Question-65753

Question Number 65753 by Masumsiddiqui399@gmail.com last updated on 03/Aug/19 Answered by mr W last updated on 04/Aug/19 $$\mathrm{1}\:\:\:^{\left.\ast\right)} \\ $$$$\sqrt{\mathrm{1}+\mathrm{1}} \\ $$$$\sqrt{\sqrt{\mathrm{1}+\mathrm{1}}+\mathrm{1}} \\ $$$$\sqrt{\sqrt{\sqrt{\mathrm{1}+\mathrm{1}}+\mathrm{1}}+\mathrm{1}} \\…

Question-131288

Question Number 131288 by mohammad17 last updated on 03/Feb/21 Answered by liberty last updated on 11/Feb/21 $$\mathrm{L}=\int\:\sqrt{\frac{\mathrm{1}−\mathrm{3x}^{−\mathrm{3}} }{\mathrm{x}^{\mathrm{8}} }}\:\mathrm{dx}\:=\:\int\:\mathrm{x}^{−\mathrm{4}} \:\sqrt{\mathrm{1}−\mathrm{3x}^{−\mathrm{3}} }\:\mathrm{dx} \\ $$$$\mathrm{change}\:\mathrm{of}\:\mathrm{variable}\::\:\sqrt{\mathrm{1}−\mathrm{3x}^{−\mathrm{3}} }\:=\:\mathrm{h}\:\mathrm{or}\:\mathrm{1}−\mathrm{3x}^{−\mathrm{3}} =\mathrm{h}^{\mathrm{2}}…

0-1-e-x-ln-x-dx-

Question Number 131283 by SEKRET last updated on 03/Feb/21 $$\:\int_{\mathrm{0}} ^{+\infty} \:\frac{\mathrm{1}}{\boldsymbol{\mathrm{e}}^{\boldsymbol{\mathrm{x}}} \centerdot\boldsymbol{\mathrm{ln}}\left(\boldsymbol{\mathrm{x}}\right)}\:\boldsymbol{\mathrm{dx}}=? \\ $$$$ \\ $$$$ \\ $$ Terms of Service Privacy Policy Contact:…