Question Number 48067 by maxmathsup by imad last updated on 18/Nov/18 $${let}\:{y}>\mathrm{0}\:{give}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{x}^{{y}} }{{e}^{{x}} −\mathrm{1}}{dx}\:{at}\:{form}\:{of}\:{series}. \\ $$ Commented by maxmathsup by imad last updated…
Question Number 113600 by eric last updated on 14/Sep/20 $${Prouver}\:{que} \\ $$$$\beta\left({a},{b}\right)=\frac{\Gamma\left({a}\right)\Gamma\left({b}\right)}{\Gamma\left({a}+{b}\right)}=\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{{a}−\mathrm{1}} \left(\mathrm{1}−{x}\right)^{{b}−\mathrm{1}} {dx} \\ $$ Answered by Dwaipayan Shikari last updated on…
Question Number 48064 by maxmathsup by imad last updated on 18/Nov/18 $${calculate}\:{A}\:=\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{1}+{x}^{\mathrm{2}} \right)\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }{dx}\:\:−\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{1}−{x}^{\mathrm{2}} \right)\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }{dx} \\ $$ Commented by maxmathsup…
Question Number 48063 by maxmathsup by imad last updated on 18/Nov/18 $${let}\:{W}\left({x}\right)\:=\int_{−\infty} ^{+\infty} \:\:\frac{{arctan}\left({xt}^{\mathrm{2}} \right)}{\mathrm{2}+{t}^{\mathrm{2}} }{dt} \\ $$$$\left.\mathrm{1}\right)\:{find}\:{a}\:{explicit}\:{form}\:{of}\:{f}\left({x}\right) \\ $$$$\left.\mathrm{2}\right)\:{find}\:{the}\:{value}\:{of}\:\:\int_{−\infty} ^{+\infty} \:\:\:\:\frac{{t}^{\mathrm{2}} }{\left(\mathrm{2}+{t}^{\mathrm{2}} \right)\left(\mathrm{1}+{x}^{\mathrm{2}} {t}^{\mathrm{4}}…
Question Number 48062 by maxmathsup by imad last updated on 18/Nov/18 $${calculate}\:\int_{−\infty} ^{+\infty} \:\:\frac{\left({x}^{\mathrm{2}} −\mathrm{3}\right){sin}\left(\mathrm{2}{x}^{\mathrm{2}} \right)}{\left({x}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{3}} }{dx} \\ $$ Terms of Service Privacy Policy…
Question Number 48057 by F_Nongue last updated on 18/Nov/18 Commented by maxmathsup by imad last updated on 18/Nov/18 $${I}\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\frac{\mathrm{1}−{x}}{\mathrm{2}−{x}\:+\mathrm{3}−{x}}{dx}\:+\int_{\mathrm{1}} ^{\mathrm{2}} \:\:\frac{{x}−\mathrm{1}}{\mathrm{2}−{x}\:+\mathrm{3}−{x}}\:+\int_{\mathrm{2}} ^{\mathrm{3}} \:\:\frac{{x}−\mathrm{1}}{{x}−\mathrm{2}\:+\mathrm{3}−{x}}\:+\int_{\mathrm{3}}…
Question Number 48042 by maxmathsup by imad last updated on 18/Nov/18 $${find}\:{the}\:{value}\:{of}\:\:\int_{−\infty} ^{+\infty} \:\:\:\frac{\mathrm{2}{x}+\mathrm{1}}{\left({x}^{\mathrm{2}} +{i}\right)\left({x}^{\mathrm{2}} +\mathrm{4}\right)}{dx}\:\:\:\left({i}^{\mathrm{2}} =−\mathrm{1}\right) \\ $$ Commented by maxmathsup by imad last…
Question Number 48043 by maxmathsup by imad last updated on 18/Nov/18 $${let}\:{f}\left(\alpha\right)\:=\int_{−\infty} ^{+\infty} \:\:\frac{{cos}\left(\alpha{x}^{\mathrm{3}} \right)}{{x}^{\mathrm{2}} \:+\mathrm{8}}\:{dx} \\ $$$$\left.\mathrm{1}\right){calculate}\:{f}\left(\alpha\right) \\ $$$$\left.\mathrm{2}\right)\:{calculate}\:\int_{−\infty} ^{+\infty} \:\:\frac{{cos}\left(\mathrm{2}{x}^{\mathrm{3}} \right)}{{x}^{\mathrm{2}} \:+\mathrm{8}}{dx}\:. \\…
Question Number 48040 by maxmathsup by imad last updated on 18/Nov/18 $${let}\:{f}\left(\alpha\right)=\int_{−\infty} ^{+\infty} \:\:\frac{{xsin}\left(\alpha{x}\right)}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\mathrm{2}} }{dx} \\ $$$${calculate}\:{f}\left(\alpha\right)\:{and}\:{f}^{'} \left(\alpha\right).\left(\alpha\:{from}\:{R}\right)\:. \\ $$ Commented by Abdo msup.…
Question Number 179107 by Acem last updated on 25/Oct/22 $${I}=\:\int\:\frac{{x}^{\mathrm{2}} }{\mathrm{sin}\:\left(\mathrm{2arctan}\:\left({e}^{{x}} \right)\right)}\:{dx}\:\:,\:{Find}\:{I} \\ $$ Answered by MJS_new last updated on 25/Oct/22 $$\int\frac{{x}^{\mathrm{2}} }{\mathrm{sun}\:\mathrm{2arctan}\:\mathrm{e}^{{x}} }{dx}=\int{x}^{\mathrm{2}} \mathrm{cosh}\:{x}\:{dx}=…