Question Number 194388 by SaRahAli last updated on 05/Jul/23 $$\int\frac{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)\mathrm{sin}\:{x}+\mathrm{4cos}\:{x}}{\mathrm{2}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\mathrm{3}} }\:{dx} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 194412 by SaRahAli last updated on 05/Jul/23 Answered by som(math1967) last updated on 05/Jul/23 $$\:\int{ln}\left({lnx}\right){dx}+\int\frac{{dx}}{\left({lnx}\right)^{\mathrm{2}} } \\ $$$$={ln}\left({lnx}\right)\int{dx}−\int\left\{\frac{{d}}{{dx}}{ln}\left({lnx}\right)\int{dx}\right\}{dx} \\ $$$$\:\:+\int\frac{{dx}}{\left({lnx}\right)^{\mathrm{2}} } \\ $$$$={xln}\left({lnx}\right)−\int\frac{{xdx}}{{xlnx}}\:+\int\frac{{dx}}{\left({lnx}\right)^{\mathrm{2}}…
Question Number 193982 by cortano12 last updated on 25/Jun/23 $$\:\:\:\:\:\int\:\frac{\mathrm{6x}^{\mathrm{3}} +\mathrm{9x}^{\mathrm{2}} +\mathrm{15x}+\mathrm{6}}{\:\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{x}+\mathrm{1}}}\:\mathrm{dx}\:=? \\ $$ Answered by MM42 last updated on 26/Jun/23 $${I}=\left({ax}^{\mathrm{2}} +{bx}+{e}\right)\sqrt{{x}^{\mathrm{2}} +{x}+\mathrm{1}}+{f}\int\frac{{dx}}{\:\sqrt{{x}^{\mathrm{2}}…
Question Number 193852 by SaRahAli last updated on 21/Jun/23 Answered by cortano12 last updated on 21/Jun/23 $$\:\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\:\int\:\frac{\mathrm{sin}\:\mathrm{2x}}{\mathrm{cos}\:\left(\mathrm{x}−\frac{\pi}{\mathrm{4}}\right)}=\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\:\int\:\frac{\mathrm{cos}\:\mathrm{2u}}{\mathrm{cos}\:\mathrm{u}}\:\mathrm{du} \\ $$$$\:=\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\:\int\:\frac{\mathrm{2cos}\:^{\mathrm{2}} \mathrm{u}−\mathrm{1}}{\mathrm{cos}\:\mathrm{u}}\:\mathrm{du} \\ $$$$\:=\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\:\left[\int\:\mathrm{2cos}\:\mathrm{u}\:\mathrm{du}−\int\mathrm{sec}\:\mathrm{u}\:\mathrm{du}\:\right] \\ $$$$\:=\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\:\left[\:\mathrm{2sin}\:\mathrm{u}−\mathrm{ln}\:\mid\mathrm{sec}\:\mathrm{u}+\mathrm{tan}\:\mathrm{u}\mid\:+\mathrm{c}\:\right. \\…
Question Number 193526 by SaRahAli last updated on 16/Jun/23 Answered by MM42 last updated on 16/Jun/23 $${I}=\int\:\frac{{sinx}×{cosx}}{{sinx}+{cos}^{\mathrm{2}} {x}}\:{dx}=−\int\:\frac{{sinx}×{cosx}}{{sinx}−{sinx}−\mathrm{1}}\:{dx} \\ $$$${let}\:\:\:{sinx}={u}\Rightarrow{cosxdx}={du} \\ $$$$\Rightarrow{I}=−\int\:\frac{{udu}}{{u}^{\mathrm{2}} −{u}−\mathrm{1}}\:{du}\:=\int\:\frac{{A}}{{u}−{u}_{\mathrm{1}} }\:{du}\:+\int\:\frac{{B}}{{u}−{u}_{\mathrm{2}} }\:{du}\:…
Question Number 193538 by cortano12 last updated on 16/Jun/23 $$\:\:\:\int\:\frac{\mathrm{dx}}{\mathrm{x}^{\mathrm{6}} +\mathrm{1}}\:=? \\ $$ Answered by Subhi last updated on 16/Jun/23 $$\int\frac{\mathrm{1}}{\left({x}^{\mathrm{2}} +\mathrm{1}\right)\left({x}^{\mathrm{4}} −{x}^{\mathrm{2}} +\mathrm{1}\right)}\:\Rrightarrow\:\int\frac{{a}}{{x}^{\mathrm{2}} +\mathrm{1}}+\frac{{bx}^{\mathrm{2}}…
Question Number 193512 by SaRahAli last updated on 15/Jun/23 Answered by Subhi last updated on 15/Jun/23 $$\int\frac{\mathrm{1}}{\frac{\mathrm{1}}{{cos}\left({x}\right)}.\frac{{sin}\left({x}\right)}{{cos}\left({x}\right)}}\:\Rrightarrow\:\int\frac{{cos}^{\mathrm{2}} \left({x}\right)}{{sin}\left({x}\right)}.{dx} \\ $$$$\int\frac{\mathrm{1}−{sin}^{\mathrm{2}} \left({x}\right)}{{sin}\left({x}\right)}.{dx}\:\:\looparrowright\:\int{csc}\left({x}\right)\:+\:\int−{sin}\left({x}\right) \\ $$$$\int{csc}\left({x}\right)\:+{cos}\left({x}\right) \\ $$$$\int{csc}\left({x}\right).\frac{{csc}\left({x}\right)+{cot}\left({x}\right)}{{csc}\left({x}\right)+{cot}\left({x}\right)}\:+\:{cos}\left({x}\right)…
Question Number 193439 by Nimnim111118 last updated on 14/Jun/23 $$\underset{\:\:\mathrm{0}} {\int}^{\pi/\mathrm{2}} \frac{\sqrt[{\mathrm{3}}]{{tanx}}}{\left({sinx}+{cosx}\right)^{\mathrm{2}} }{dx} \\ $$ Commented by Nimnim111118 last updated on 14/Jun/23 $${help}… \\ $$…
Question Number 193409 by cortano12 last updated on 13/Jun/23 $$\:\:\:\Subset \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 193377 by Humble last updated on 12/Jun/23 $${Evaluate} \\ $$$${I}=\int_{\mathrm{0}} ^{\:\infty} \frac{\mathrm{1}}{{x}^{\mathrm{5}} +{x}^{\mathrm{4}} +{x}^{\mathrm{3}} +{x}^{\mathrm{2}} +{x}+\mathrm{1}}{dx} \\ $$ Commented by Frix last updated…