Question Number 40657 by Raj Singh last updated on 25/Jul/18 Answered by math khazana by abdo last updated on 25/Jul/18 $${let}\:{I}\:=\:\int\:\frac{{a}+{x}}{{a}−{x}}{dx} \\ $$$${I}\:=\:\int\:\frac{{a}−{x}+\mathrm{2}{x}}{{a}−{x}}{dx}\:=\int\:{dx}\:\:+\mathrm{2}\:\int\:\frac{{x}}{{a}−{x}}{dx} \\ $$$$={x}\:−\mathrm{2}\:\int\:\frac{{a}−{x}−{a}}{{a}−{x}}{dx}=\:{x}−\mathrm{2}\:\int\:{dx}\:\:+\mathrm{2}{a}\:\int\:\:\frac{{dx}}{{a}−{x}}…
Question Number 40658 by math khazana by abdo last updated on 25/Jul/18 $${find}\:\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:\frac{{x}−\mathrm{1}}{\mathrm{2}+{cosx}}{dx}\:. \\ $$ Commented by math khazana by abdo last updated…
Question Number 171727 by cortano1 last updated on 20/Jun/22 Answered by mahdipoor last updated on 20/Jun/22 $${y}\geqslant\frac{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }{\mathrm{2}}\Rightarrow\mathrm{1}\geqslant{x}^{\mathrm{2}} +\left({y}−\mathrm{1}\right)^{\mathrm{2}} \Rightarrow\:−\mathrm{1}\leqslant{x}\leqslant\mathrm{1} \\ $$$${for}\:{x}={x}_{\mathrm{0}} \:,\:{max}\left({x}_{\mathrm{0}} +{y}\right)={x}_{\mathrm{0}}…
Question Number 171708 by ilhamQ last updated on 20/Jun/22 $$\int_{\mathrm{0}} ^{\infty} \:\mathrm{2}{x}−\mathrm{3}\:{dx}=… \\ $$ Answered by puissant last updated on 20/Jun/22 $$=\:\underset{{A}\rightarrow\infty} {\mathrm{lim}}\:\int_{\mathrm{0}} ^{{A}} \mathrm{2}{x}−\mathrm{3}\:{dx}…
Question Number 40625 by Raj Singh last updated on 25/Jul/18 Answered by MJS last updated on 25/Jul/18 $$\int\frac{{dx}}{\:\sqrt{{x}}+\sqrt[{\mathrm{3}}]{{x}}}= \\ $$$$\:\:\:\:\:\left[{t}=\sqrt[{\mathrm{6}}]{{x}}\:\rightarrow\:{dx}=\mathrm{6}\sqrt[{\mathrm{6}}]{{x}^{\mathrm{5}} }{dt}\right] \\ $$$$=\mathrm{6}\int\frac{{t}^{\mathrm{3}} }{{t}+\mathrm{1}}{dt}=\mathrm{6}\int\left({t}^{\mathrm{2}} −{t}+\mathrm{1}−\frac{\mathrm{1}}{{t}+\mathrm{1}}\right){dt}=…
Question Number 40624 by math khazana by abdo last updated on 25/Jul/18 $${let}\:{f}\left({x}\right)=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {ln}\left(\frac{\mathrm{1}−{xsint}}{\mathrm{1}+{xsint}}\right){dt}\:\:. \\ $$$$\left.\mathrm{1}\right)\:{find}\:{the}\:{value}\:{of}\:\:{I}\:=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:{ln}\left(\mathrm{1}−{xsint}\right){dt} \\ $$$${and}\:{J}\:=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {ln}\left(\mathrm{1}+{xsint}\right){dt} \\ $$$$\left.\mathrm{2}\right)\:{find}\:{a}\:{simple}\:{form}\:{of}\:{f}\left({x}\right)…
let-f-x-0-pi-2-ln-1-xcos-d-1-calculate-f-1-2-find-a-simple-form-of-f-x-3-developp-f-at-ontehr-serie-
Question Number 40621 by math khazana by abdo last updated on 25/Jul/18 $${let}\:{f}\left({x}\right)=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {ln}\left(\mathrm{1}+{xcos}\theta\right){d}\theta\: \\ $$$$\left.\mathrm{1}\right)\:{calculate}\:{f}\left(\mathrm{1}\right) \\ $$$$\left.\mathrm{2}\right)\:{find}\:{a}\:{simple}\:{form}\:{of}\:{f}\left({x}\right) \\ $$$$\left.\mathrm{3}\right)\:{developp}\:{f}\:{at}\:{ontehr}\:{serie} \\ $$ Answered by…
Question Number 40619 by math khazana by abdo last updated on 25/Jul/18 $${let}\:\:{f}\left({x}\right)=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\:\frac{{d}\theta}{{x}\:\:+{cos}^{\mathrm{2}} \theta}\:\:{with}\:{x}>\mathrm{0}\:. \\ $$$$\left.\mathrm{1}\right)\:{calculate}\:{f}\left({x}\right)\:{and}\:{f}^{'} \left({x}\right) \\ $$$$\left.\mathrm{2}\right)\:{find}\:\:{f}^{\left({n}\right)} \left({x}\right)\:{and}\:{f}^{\left({n}\right)} \left(\mathrm{0}\right) \\ $$$$\left.\mathrm{3}\right)\:{developp}\:{f}\:{at}\:{integr}\:{serie}.…
Question Number 40620 by math khazana by abdo last updated on 25/Jul/18 $${find}\:\:\int\:\:\:\left(\left({x}+\mathrm{1}\right)\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\:+\left(\mathrm{1}+{x}^{\mathrm{2}} \right)\sqrt{{x}+\mathrm{1}}\right){dx} \\ $$ Answered by tanmay.chaudhury50@gmail.com last updated on 25/Jul/18 $$\int{x}\sqrt{\mathrm{1}+{x}^{\mathrm{2}}…
Question Number 106148 by mohammad17 last updated on 03/Aug/20 $$\left(\mathrm{1}\right)\int\:{sec}^{−\mathrm{1}} {x}^{\mathrm{2}} {dx} \\ $$$$ \\ $$$$\left(\mathrm{2}\right)\:\int\sqrt{{x}+\mathrm{2}}\:\:{sin}^{−\mathrm{1}} \sqrt{\mathrm{3}{x}−\mathrm{1}}\:{dx} \\ $$ Answered by bemath last updated on…