Question Number 120040 by bramlexs22 last updated on 28/Oct/20 $$\:\int\:\frac{{t}^{\mathrm{5}} }{\:\sqrt{\mathrm{2}+{t}^{\mathrm{2}} }}\:{dt}\: \\ $$ Answered by john santu last updated on 28/Oct/20 $$\:{let}\:\mathrm{2}+{t}^{\mathrm{2}} \:=\:{h}^{\mathrm{2}} \:\rightarrow\:{t}\:{dt}\:=\:{h}\:{dh}\:…
Question Number 120025 by mathmax by abdo last updated on 28/Oct/20 $$\mathrm{calculate}\:\mathrm{f}^{'} \left(\mathrm{x}\right) \\ $$$$\left.\mathrm{1}\right)\:\mathrm{f}\left(\mathrm{x}\right)\:=\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{cos}\left(\mathrm{xt}\right)}{\mathrm{t}^{\mathrm{2}} +\mathrm{x}^{\mathrm{2}} }\mathrm{dt} \\ $$$$\left.\mathrm{2}\right)\mathrm{f}\left(\mathrm{x}\right)=\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{sin}\left(\mathrm{xt}^{\mathrm{2}} +\sqrt{\mathrm{2}}\right)}{\mathrm{t}^{\mathrm{2}} +\mathrm{x}^{\mathrm{2}}…
Question Number 119989 by mnjuly1970 last updated on 28/Oct/20 Answered by mathmax by abdo last updated on 28/Oct/20 $$\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{xdx}}{\mathrm{2e}^{\mathrm{x}} −\mathrm{1}}\:=\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{xe}^{−\mathrm{x}} \mathrm{dx}}{\mathrm{2}−\mathrm{e}^{−\mathrm{x}}…
Question Number 119970 by bramlexs22 last updated on 28/Oct/20 $$\:\int\:\frac{{dx}}{{x}^{\mathrm{2}} \sqrt{\mathrm{25}−{x}^{\mathrm{2}} }}\:? \\ $$ Answered by bemath last updated on 28/Oct/20 $$\:\int\:\frac{{dx}}{{x}^{\mathrm{3}} \sqrt{\frac{\mathrm{25}}{{x}^{\mathrm{2}} }−\mathrm{1}}}\:{dx}\:=\:\int\:\frac{{x}^{−\mathrm{3}} }{\:\sqrt{\mathrm{25}{x}^{−\mathrm{2}}…
Question Number 119960 by mnjuly1970 last updated on 28/Oct/20 Answered by mathmax by abdo last updated on 28/Oct/20 $$\mathrm{A}\:=\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{arctanx}}{\left(\mathrm{1}+\mathrm{x}\right)\sqrt{\mathrm{x}}}\mathrm{dx}\:\Rightarrow\mathrm{A}\:=_{\sqrt{\mathrm{x}}=\mathrm{t}} \:\:\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{arctan}\left(\mathrm{t}^{\mathrm{2}} \right)}{\left(\mathrm{1}+\mathrm{t}^{\mathrm{2}}…
Question Number 54416 by 951172235v last updated on 03/Feb/19 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 185484 by Mingma last updated on 22/Jan/23 Answered by Ar Brandon last updated on 22/Jan/23 $${I}=\int_{\mathrm{0}} ^{\infty} \frac{{x}^{\mathrm{3}} }{{e}^{\mathrm{3}{x}} −\mathrm{1}}{dx}=\int_{\mathrm{0}} ^{\infty} \frac{{x}^{\mathrm{3}} {e}^{−\mathrm{3}{x}}…
Question Number 119934 by bramlexs22 last updated on 28/Oct/20 $$\:\:\:\int\:\frac{\mathrm{sin}\:^{\mathrm{8}} {x}−\mathrm{cos}\:^{\mathrm{8}} {x}}{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sin}\:^{\mathrm{2}} \mathrm{2}{x}}\:{dx}\: \\ $$ Answered by bobhans last updated on 28/Oct/20 $$\:\int\:\frac{\left(\mathrm{sin}\:^{\mathrm{4}} {x}−\mathrm{cos}\:^{\mathrm{4}} {x}\right)\left(\mathrm{sin}\:^{\mathrm{4}}…
Question Number 119930 by Lordose last updated on 28/Oct/20 Answered by mathmax by abdo last updated on 28/Oct/20 $$\mathrm{A}\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:\mathrm{arcsin}\left(\mathrm{tan}\theta\right)\mathrm{d}\theta\:\:\mathrm{we}\:\mathrm{do}\:\mathrm{the}\:\mathrm{changement} \\ $$$$\mathrm{arcsin}\left(\mathrm{tan}\theta\right)=\mathrm{x}\:\Rightarrow\mathrm{tan}\theta\:=\mathrm{sinx}\:\Rightarrow\theta\:=\mathrm{artan}\left(\mathrm{sinx}\right)\:\Rightarrow \\ $$$$\mathrm{A}\:=\int_{\mathrm{0}}…
Question Number 119920 by bramlexs22 last updated on 28/Oct/20 $$\:\int\:\sqrt{\frac{\mathrm{1}−{x}}{\mathrm{1}+{x}}}\:{dx}\:=\:?\:,\:{x}\epsilon\left(−\mathrm{1},\mathrm{1}\right) \\ $$ Answered by mathmax by abdo last updated on 28/Oct/20 $$\mathrm{I}\:=\int\sqrt{\frac{\mathrm{1}−\mathrm{x}}{\mathrm{1}+\mathrm{x}}}\mathrm{dx}\:\:\mathrm{we}\:\mathrm{do}\:\mathrm{the}\:\mathrm{changement}\:\mathrm{x}\:=\mathrm{cost}\:\Rightarrow \\ $$$$\mathrm{I}\:=\int\:\sqrt{\frac{\mathrm{2sin}^{\mathrm{2}} \left(\frac{\mathrm{t}}{\mathrm{2}}\right)}{\mathrm{2cos}^{\mathrm{2}}…