Question Number 99239 by abdomathmax last updated on 19/Jun/20 $$\mathrm{calculate}\:\int_{\mathrm{2}} ^{+\infty} \:\frac{\mathrm{dx}}{\mathrm{x}^{\mathrm{3}} \left(\mathrm{x}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} } \\ $$ Answered by MJS last updated on 19/Jun/20 $$\mathrm{Ostrogradski}\:\mathrm{gives}…
Question Number 99234 by abdomathmax last updated on 19/Jun/20 $$\mathrm{calculate}\:\int_{\mathrm{0}} ^{\infty} \:\mathrm{e}^{−\mathrm{x}} \mathrm{lnx}\:\mathrm{dx} \\ $$ Answered by maths mind last updated on 19/Jun/20 $$\Gamma\left({x}\right)=\int_{\mathrm{0}} ^{\infty}…
Question Number 33695 by math khazana by abdo last updated on 22/Apr/18 $${find}\:{lim}_{{n}\rightarrow+\infty} \:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\:\frac{{e}^{−\frac{{x}}{{n}}} }{\mathrm{1}+{x}^{\mathrm{2}} }{dx}. \\ $$ Commented by math khazana by…
Question Number 99228 by M±th+et+s last updated on 19/Jun/20 $$ \\ $$$$\int_{\mathrm{0}} ^{\infty} \frac{{sin}\left({x}\right){ln}\left({x}\right)}{{x}}{dx}=\frac{−\gamma\pi}{\mathrm{2}} \\ $$ Answered by maths mind last updated on 20/Jun/20 $$\frac{{ln}\left({x}\right)}{{x}}=\frac{\partial}{\partial{a}}{x}^{{a}−\mathrm{1}}…
Question Number 33694 by math khazana by abdo last updated on 22/Apr/18 $${calculate}\:{lim}_{{n}\rightarrow+\infty} \:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{dx}}{{x}^{{n}} \:\:+{e}^{{x}} }\:\:. \\ $$ Commented by math khazana by…
Question Number 33689 by NECx last updated on 22/Apr/18 $$\int\frac{{x}}{{x}^{\mathrm{3}} +\mathrm{1}}{dx} \\ $$ Commented by mondodotto@gmail.com last updated on 22/Apr/18 $$\boldsymbol{\mathrm{let}}\:\boldsymbol{\mathrm{u}}=\boldsymbol{\mathrm{x}}^{\mathrm{3}} +\mathrm{1} \\ $$$$\frac{\boldsymbol{\mathrm{du}}}{\boldsymbol{\mathrm{dx}}}=\mathrm{3}\boldsymbol{\mathrm{x}}^{\mathrm{2}} \Rightarrow\boldsymbol{\mathrm{dx}}=\frac{\boldsymbol{\mathrm{du}}}{\mathrm{3}\boldsymbol{\mathrm{x}}^{\mathrm{2}}…
Question Number 33677 by math khazana by abdo last updated on 21/Apr/18 $${calculate}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{xlnx}}{{x}−\mathrm{1}}{dx}\:. \\ $$ Commented by math khazana by abdo last updated…
Question Number 99205 by bemath last updated on 19/Jun/20 Commented by bramlex last updated on 19/Jun/20 $${let}\:\sqrt{\mathrm{4}−\sqrt{{x}}}\:=\:{t}\:,\:\mathrm{4}−\sqrt{{x}}\:=\:{t}^{\mathrm{2}} \\ $$$$\left(\mathrm{4}−{t}^{\mathrm{2}} \right)^{\mathrm{2}} \:=\:{x}\:;\:{dx}\:=\:−\mathrm{4}{t}\left(\mathrm{4}−{t}^{\mathrm{2}} \right)\:{dt}\: \\ $$$$\mathrm{I}\:=\:\int\:{t}\:\left(−\mathrm{4}{t}\left(\mathrm{4}−{t}^{\mathrm{2}} \right)\right)\:{dt}\:…
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Question Number 99173 by bemath last updated on 19/Jun/20 Answered by MJS last updated on 19/Jun/20 $$\mathrm{this}\:\mathrm{is}\:\mathrm{very}\:\mathrm{complicated} \\ $$$$\int\sqrt{\frac{\mathrm{1}+\mathrm{cot}\:{x}}{\mathrm{csc}\:{x}\:+\mathrm{cot}\:{x}}}{dx}=\int\sqrt{\frac{\mathrm{sin}\:{x}\:+\mathrm{cos}\:{x}}{\mathrm{1}+\mathrm{cos}\:{x}}}{dx}= \\ $$$$\:\:\:\:\:\left[{t}=\mathrm{tan}\:\frac{{x}}{\mathrm{2}}\:\rightarrow\:{dx}=\mathrm{2cos}^{\mathrm{2}} \:\frac{{x}}{\mathrm{2}}\:{dt}\right] \\ $$$$=\sqrt{\mathrm{2}}\int\frac{\sqrt{−{t}^{\mathrm{2}} +\mathrm{2}{t}+\mathrm{1}}}{{t}^{\mathrm{2}}…