Question Number 108710 by mathmax by abdo last updated on 18/Aug/20 $$\mathrm{calculate}\:\:\int_{−\infty} ^{\infty} \:\:\frac{\left(−\mathrm{1}\right)^{\mathrm{x}^{\mathrm{2}} } }{\left(\mathrm{x}^{\mathrm{2}} +\mathrm{x}+\mathrm{1}\right)^{\mathrm{2}} }\mathrm{dx} \\ $$ Commented by sunilbaraskar last updated…
Question Number 43170 by maxmathsup by imad last updated on 07/Sep/18 $${let}\:{A}_{{n}} =\:\int_{\mathrm{0}} ^{\infty} \:\:\left[{n}\:{e}^{−{x}} \right]{dx}\:\:{with}\:{n}\:{integr}\:{natural}. \\ $$$$\left.\mathrm{1}\right)\:{calculate}\:{A}_{{n}} \\ $$$$\left.\mathrm{2}\right)\:{find}\:{lim}_{{n}\rightarrow+\infty} \:{A}_{{n}} . \\ $$ Commented…
Question Number 108697 by 150505R last updated on 18/Aug/20 Answered by mathmax by abdo last updated on 18/Aug/20 $$\mathrm{I}\:=\int_{\mathrm{0}} ^{\infty} \:\:\frac{\left(\mathrm{lnx}\right)^{\mathrm{2}} }{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }\:\mathrm{dx}\:\Rightarrow\:\mathrm{I}\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\left(\mathrm{lnx}\right)^{\mathrm{2}}…
Question Number 43158 by MASANJA J last updated on 07/Sep/18 $$\int{cosecxdx} \\ $$ Commented by maxmathsup by imad last updated on 07/Sep/18 $${let}\:{I}\:=\:\int\:\:\:\frac{{dx}}{{sin}\left({x}\right)}\:{changement}\:{tan}\left(\frac{{x}}{\mathrm{2}}\right)={t}\:{give} \\ $$$${I}\:=\:\int\:\:\:\:\frac{\mathrm{1}}{\frac{\mathrm{2}{t}}{\mathrm{1}+{t}^{\mathrm{2}}…
Question Number 43159 by MASANJA J last updated on 07/Sep/18 Answered by alex041103 last updated on 08/Sep/18 $${For}\:\int\frac{{dx}}{\:\sqrt{{x}+\mathrm{15}}}\:: \\ $$$$\int\frac{{dx}}{\:\sqrt{{x}+\mathrm{15}}}\:=\:\int\frac{{d}\left({x}+\mathrm{15}\right)}{\:\sqrt{{x}+\mathrm{15}}}=\int{u}^{−\mathrm{1}/\mathrm{2}} {du}= \\ $$$$=\mathrm{2}{u}^{\mathrm{1}/\mathrm{2}} +{C}=\mathrm{2}\sqrt{{x}+\mathrm{15}}\:+{C} \\…
Question Number 108692 by 150505R last updated on 18/Aug/20 Answered by Dwaipayan Shikari last updated on 18/Aug/20 $$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\mathrm{2}{log}\left({tan}\theta\right)}{{tan}^{\mathrm{2}} \theta+\mathrm{1}}{sec}^{\mathrm{2}} \theta{d}\theta\:\:\:\left({x}={tan}\theta,\:\mathrm{1}={sec}^{\mathrm{2}} \theta\frac{{d}\theta}{{dx}}\right) \\ $$$$\mathrm{2}\int_{\mathrm{0}}…
Question Number 43156 by MASANJA J last updated on 07/Sep/18 $${integrate}\:{w}.{r}.{t}\:{x} \\ $$$$\int\frac{{xe}^{{x}} }{\:\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}{dx} \\ $$ Answered by tanmay.chaudhury50@gmail.com last updated on 09/Sep/18 $${x}={tan}\alpha\:\:{dx}={sec}^{\mathrm{2}}…
Question Number 43157 by MASANJA J last updated on 07/Sep/18 $$\int{secxdx} \\ $$ Commented by maxmathsup by imad last updated on 07/Sep/18 $${let}\:{J}\:=\:\int\:\:\frac{{dx}}{{cosx}}\:{changement}\:{tan}\left(\frac{{x}}{\mathrm{2}}\right)={u}\:{give} \\ $$$${J}\:=\:\int\:\:\frac{\mathrm{1}}{\frac{\mathrm{1}−{u}^{\mathrm{2}}…
Question Number 43147 by rahul 19 last updated on 07/Sep/18 $$\mathrm{If}\:\:\:\int_{\mathrm{0}} ^{\infty} \:\mathrm{e}^{−{x}^{\mathrm{2}} } {dx}\:=\:\frac{\sqrt{\pi}}{\mathrm{2}\:}\:, \\ $$$$\mathrm{then}\:{prove}\:{that}\:\int_{\mathrm{0}} ^{\infty} \mathrm{e}^{−\mathrm{a}{x}^{\mathrm{2}} } {dx}\:=\:\sqrt{\frac{\pi}{\mathrm{4a}}} \\ $$$$\mathrm{where}\:\mathrm{a}>\mathrm{0}. \\ $$…
Question Number 43145 by rahul 19 last updated on 07/Sep/18 $$\int_{\mathrm{0}} ^{\infty} \:\left[\:\mathrm{2e}^{−{x}} \right]{dx}\:=\:?\: \\ $$$${where}\:\left[.\right]=\:{gif}. \\ $$ Commented by maxmathsup by imad last updated…