Question Number 191022 by Rupesh123 last updated on 16/Apr/23 Commented by Frix last updated on 16/Apr/23 $$\mathrm{1}? \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 125482 by joki last updated on 11/Dec/20 Answered by MJS_new last updated on 11/Dec/20 $$\int\frac{\mathrm{4}{x}^{\mathrm{3}} +{x}^{\mathrm{2}} +\mathrm{1}}{\left({x}^{\mathrm{2}} −\mathrm{2}\right)^{\mathrm{3}} }{dx}= \\ $$$$\:\:\:\:\:\left[\mathrm{Ostrogradski}'\mathrm{s}\:\mathrm{Method}\right] \\ $$$$=\frac{{x}^{\mathrm{3}}…
Question Number 59946 by Sardor2211 last updated on 16/May/19 Commented by maxmathsup by imad last updated on 17/May/19 $${let}\:{use}\:{the}\:{chang}\:\:\mathrm{1}+\left({x}+\mathrm{1}\right)^{\frac{\mathrm{1}}{\mathrm{3}}} \:={t}\:\Rightarrow\left({x}+\mathrm{1}\right)^{\frac{\mathrm{1}}{\mathrm{3}}} \:={t}−\mathrm{1}\:\Rightarrow{x}+\mathrm{1}\:=\left({t}−\mathrm{1}\right)^{\mathrm{3}} \:\Rightarrow \\ $$$$\int_{−\mathrm{1}} ^{\mathrm{0}}…
Question Number 191019 by Albert12 last updated on 16/Apr/23 $$\mathrm{1}−\frac{\mathrm{1}−\frac{\mathrm{1}−\frac{\iddots}{\mathrm{5}}}{\mathrm{5}}}{\mathrm{5}}\:\:=? \\ $$ Answered by 073 last updated on 16/Apr/23 $$\mathrm{1}−\frac{\mathrm{x}}{\mathrm{5}}=\mathrm{x} \\ $$$$\mathrm{1}=\mathrm{x}+\frac{\mathrm{x}}{\mathrm{5}}=\frac{\mathrm{6x}}{\mathrm{5}} \\ $$$$\mathrm{6x}=\mathrm{5} \\…
Question Number 59942 by naka3546 last updated on 16/May/19 $${f}\left(\mathrm{2}{x}−\mathrm{5}\right)\:+\:{f}\left(\mathrm{3}{x}+\mathrm{5}\right)\:\:=\:\:\mathrm{7}{x}−\mathrm{16} \\ $$$${f}\left({x}\right)\:\:=\:\:? \\ $$ Answered by mr W last updated on 16/May/19 $${let}\:{f}\left({x}\right)={ax}+{b} \\ $$$${f}\left(\mathrm{2}{x}−\mathrm{5}\right)+{f}\left(\mathrm{3}{x}+\mathrm{5}\right)={a}\left(\mathrm{2}{x}−\mathrm{5}\right)+{b}+{a}\left(\mathrm{3}{x}+\mathrm{5}\right)+{b}…
Question Number 191009 by pascal889 last updated on 16/Apr/23 Answered by Frix last updated on 16/Apr/23 $$\int\frac{\mathrm{3}{x}+\mathrm{2}}{{x}^{\mathrm{2}} +\mathrm{6}{x}+\mathrm{4}}{dt}=\frac{\mathrm{3}{x}+\mathrm{2}}{{x}^{\mathrm{2}} +\mathrm{6}{x}+\mathrm{4}}\int{dt}=\frac{\mathrm{3}{x}+\mathrm{2}}{{x}^{\mathrm{2}} +\mathrm{6}{x}+\mathrm{4}}{t}+{C} \\ $$ Commented by ARUNG_Brandon_MBU…
Question Number 59936 by Sardor2211 last updated on 16/May/19 Answered by tanmay last updated on 16/May/19 $${x}\sqrt{\mathrm{1}+{y}^{\mathrm{2}} }\:+\frac{{dy}}{{dx}}.{y}.\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\:=\mathrm{0} \\ $$$${x}\sqrt{\mathrm{1}+{y}^{\mathrm{2}} }\:{dx}+{y}\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\:{dy}=\mathrm{0} \\ $$$$\frac{{xdx}}{\:\sqrt{\mathrm{1}+{x}^{\mathrm{2}}…
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Question Number 125470 by hatakekakashi1729gmailcom last updated on 11/Dec/20 $$\int_{\mathrm{0}} ^{\:\mathrm{1}} \left({x}^{{x}} \right)^{\left({x}^{{x}} \right)^{\iddots} } {dx}\:\neq\:\frac{\pi^{\mathrm{2}} }{\mathrm{12}} \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 59935 by maxmathsup by imad last updated on 16/May/19 $${sir}\:{malwan}\:{you}\:{must}\:{revise}\:\:{analytical}\:{function}\:{and}\:{complex}\:{analysis}… \\ $$ Commented by malwaan last updated on 16/May/19 $$\mathrm{O}.\mathrm{K}.\:{Sir} \\ $$ Terms…