Question Number 169864 by mathlove last updated on 11/May/22 Answered by Mathspace last updated on 11/May/22 $$\left(^{\mathrm{3}} \sqrt{{x}}\right)={t}\:\Rightarrow{x}={t}^{\mathrm{3}} \:\Rightarrow \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \left(…\right)^{\left(…\right)} {dx}=\int_{\mathrm{0}} ^{\mathrm{1}}…
Question Number 104312 by Ar Brandon last updated on 20/Jul/20 $$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \frac{\sqrt{\mathrm{sin}^{\mathrm{2}} \theta+\mathrm{2}}}{\mathrm{sin}\theta}\mathrm{d}\theta \\ $$ Answered by mathmax by abdo last updated on 21/Jul/20…
Question Number 169825 by mathlove last updated on 10/May/22 Answered by Mathspace last updated on 10/May/22 $${I}=_{{x}=\frac{\mathrm{1}}{{t}}} \:\:\:−\int_{\mathrm{0}} ^{\infty} \:\:\frac{{ln}\left(\frac{\mathrm{1}+\frac{\mathrm{1}}{{t}^{\mathrm{11}} }}{\mathrm{1}+\frac{\mathrm{1}}{{t}^{\mathrm{3}} }}\right)}{\left(\mathrm{1}+\frac{\mathrm{1}}{{t}^{\mathrm{2}} }\right)\left(−{lnt}\right)}\left(−\frac{{dt}}{{t}^{\mathrm{2}} }\right) \\…
Question Number 38746 by MJS last updated on 29/Jun/18 $$\mathrm{this}\:\mathrm{is}\:\mathrm{still}\:\mathrm{waiting}\:\mathrm{to}\:\mathrm{be}\:\mathrm{solved}… \\ $$$$\int\frac{\sqrt{\left({t}−\mathrm{1}\right){t}\left({t}+\mathrm{1}\right)}}{\mathrm{3}{t}^{\mathrm{2}} −\mathrm{4}}{dt}=? \\ $$ Commented by behi83417@gmail.com last updated on 29/Jun/18 $${this}\:{can}\:{not}\:{be}\:{difined}\:\:{in}\:{terms}\:{of} \\ $$$${primary}\:{functions}.{dont}\:{spend}\:{time}\:{on}…
Question Number 104264 by mohammad17 last updated on 20/Jul/20 $$\int_{\mathrm{0}} ^{\:\sqrt{\mathrm{2}}} \:\int_{{y}} ^{\:\sqrt{\mathrm{4}−{y}^{\mathrm{2}} }} \frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }}{dxdy} \\ $$ Answered by Ar Brandon last updated…
Question Number 38728 by maxmathsup by imad last updated on 28/Jun/18 $${find}\:{L}\:\left(\:\frac{{e}^{−\frac{{x}}{{a}}} }{{a}}\right)\:\:{with}\:{a}\neq\mathrm{0}\:\:{and}\:{L}\:{laplace}\:{transfom}. \\ $$ Commented by abdo mathsup 649 cc last updated on 29/Jun/18…
Question Number 38727 by maxmathsup by imad last updated on 28/Jun/18 $${let}\:{n}\:{from}\:{N}\:\:{and}\:\:{A}_{{n}} =\:\int_{−\infty} ^{+\infty} \:\:\:\:\frac{{cos}\left({ax}\right)}{\left({x}^{\mathrm{2}} \:+{x}+\mathrm{1}\right)^{{n}} }{dx}\:\:{and} \\ $$$${B}_{{n}} =\:\int_{−\infty} ^{+\infty} \:\:\:\frac{{sin}\left({ax}\right)}{\left({x}^{\mathrm{2}} \:+{x}+\mathrm{1}\right)^{{n}} }{dx} \\…
Question Number 38724 by maxmathsup by imad last updated on 28/Jun/18 $${calculate}\:\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{x}^{\mathrm{2}} {cos}\left(\pi{x}\right)}{\left({x}^{\mathrm{2}} \:+\mathrm{4}\right)^{\mathrm{2}} }{dx} \\ $$ Commented by maxmathsup by imad last…
Question Number 38719 by maxmathsup by imad last updated on 28/Jun/18 $${find}\:\:\:\int\:\:{ln}\left(\sqrt{{x}}\:+\sqrt{{x}+\mathrm{1}}\right){dx} \\ $$ Answered by behi83417@gmail.com last updated on 29/Jun/18 $${I}={xln}\left(\sqrt{{x}}+\sqrt{{x}+\mathrm{1}}\right)−\int{x}\frac{\frac{\mathrm{1}}{\mathrm{2}\sqrt{{x}}}+\frac{\mathrm{1}}{\mathrm{2}\sqrt{{x}+\mathrm{1}}}}{\:\sqrt{{x}}+\sqrt{{x}+\mathrm{1}}}{dx}= \\ $$$$={do}−\int\frac{{x}}{\mathrm{2}\sqrt{{x}}\sqrt{{x}+\mathrm{1}}}{dx}={do}−\int\frac{\sqrt{{x}}}{\mathrm{2}\sqrt{{x}+\mathrm{1}}}{dx} \\…
Question Number 38720 by maxmathsup by imad last updated on 28/Jun/18 $${find}\:\:\:\int\:\:\:\:\:\frac{\sqrt{{x}+\mathrm{1}}\:−\sqrt{{x}−\mathrm{1}}}{\:\sqrt{{x}+\mathrm{1}}\:−\sqrt{{x}−\mathrm{1}}}{dx} \\ $$ Commented by math khazana by abdo last updated on 28/Jun/18 $${the}\:{Q}\:{is}\:{find}\:\:\int\:\:\:\:\frac{\sqrt{{x}+\mathrm{1}}\:−\sqrt{{x}−\mathrm{1}}}{\:\sqrt{{x}+\mathrm{1}}\:+\sqrt{{x}−\mathrm{1}}}{dx}…