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Category: Integration

e-2dx-xlnx-

Question Number 84879 by sahnaz last updated on 17/Mar/20 $$\mathrm{e}^{\int\frac{\mathrm{2dx}}{\mathrm{xlnx}}} \\ $$ Commented by jagoll last updated on 17/Mar/20 $$\int\:\frac{\mathrm{2dx}}{\mathrm{x}\:\mathrm{lnx}}\:=\:\int\:\frac{\mathrm{2d}\left(\mathrm{lnx}\right)}{\mathrm{lnx}}\:=\:\int\:\mathrm{2}\frac{\mathrm{du}}{\mathrm{u}} \\ $$$$=\:\mathrm{2}\:\mathrm{ln}\:\mathrm{u}\:+\:\mathrm{c}\:,\:\left[\mathrm{u}\:=\:\mathrm{ln}\:\mathrm{x}\:\right] \\ $$$$=\:\mathrm{2ln}\left(\mathrm{lnx}\right)\:+\:\mathrm{2lnC}\:=\:\mathrm{2ln}\left(\mathrm{Clnx}\right) \\…

sin-7x-cos-3x-dx-

Question Number 84843 by M±th+et£s last updated on 16/Mar/20 $$\int\frac{{sin}\left(\mathrm{7}{x}\right)}{{cos}\left(\mathrm{3}{x}\right)}\:{dx} \\ $$ Commented by jagoll last updated on 17/Mar/20 $$\mathrm{sin}\:\mathrm{7x}\:=\:\mathrm{sin}\:\left(\mathrm{4x}+\mathrm{3x}\right)\: \\ $$$$=\:\mathrm{sin}\:\mathrm{4x}\:\mathrm{cos}\:\mathrm{3x}\:+\:\mathrm{cos}\:\mathrm{4x}\:\mathrm{sin}\:\mathrm{3x} \\ $$$$\int\:\frac{\mathrm{sin}\:\mathrm{7x}}{\mathrm{cos}\:\mathrm{3x}}\:\mathrm{dx}\:=\:\int\:\left(\mathrm{sin}\:\mathrm{4x}\:+\:\mathrm{cos}\:\mathrm{4x}\:\mathrm{tan}\:\mathrm{3x}\right)\mathrm{dx} \\…

x-x-2-1-3-2-arctan-x-dx-

Question Number 84809 by M±th+et£s last updated on 16/Mar/20 $$\int\frac{{x}}{\left({x}^{\mathrm{2}} +\mathrm{1}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} {arctan}\left({x}\right)}\:{dx} \\ $$ Commented by abdomathmax last updated on 18/Mar/20 $${A}\:=\int\:\:\:\:\frac{{x}}{\left({x}^{\mathrm{2}\:} +\mathrm{1}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} \:{arctanx}}\:{changement}\:{arctanx}={t} \\…

dx-16-9sin-x-2-

Question Number 84766 by jagoll last updated on 15/Mar/20 $$\int\:\frac{\mathrm{dx}}{\left(\mathrm{16}+\mathrm{9sin}\:\mathrm{x}\right)^{\mathrm{2}} } \\ $$$$ \\ $$ Commented by jagoll last updated on 16/Mar/20 $$\int\:\mathrm{sec}\:\:\mathrm{x}\:\left[\:\frac{\mathrm{cos}\:\:\mathrm{x}}{\left(\mathrm{16}+\mathrm{9sin}\:\mathrm{x}\right)^{\mathrm{2}} }\right]\:\mathrm{dx}\:= \\…

Question-19230

Question Number 19230 by tawa tawa last updated on 07/Aug/17 Answered by ajfour last updated on 07/Aug/17 $$\mathrm{let}\:\mathrm{g}\left(\mathrm{x}\right)=\sqrt[{\mathrm{3}}]{\mathrm{1}+\mathrm{f}\left(\mathrm{x}\right)\sqrt[{\mathrm{3}}]{\mathrm{1}+\mathrm{f}\left(\mathrm{x}−\mathrm{1}\right)\sqrt[{\mathrm{3}}]{\mathrm{1}+\mathrm{f}\left(\mathrm{x}−\mathrm{2}\right)\sqrt[{\mathrm{3}}]{\mathrm{1}+…}}}}\: \\ $$$$\Rightarrow\:\left[\mathrm{g}\left(\mathrm{x}\right)\right]^{\mathrm{3}} =\mathrm{1}+\mathrm{f}\left(\mathrm{x}\right)\mathrm{g}\left(\mathrm{x}−\mathrm{1}\right) \\ $$$$\Rightarrow\:\mathrm{degree}\:\mathrm{of}\:\mathrm{g}\left(\mathrm{x}\right)\:\mathrm{is}\:\mathrm{1}. \\ $$$$\mathrm{let}\:\mathrm{g}\left(\mathrm{x}\right)=\mathrm{Ax}+\mathrm{B}…