Question Number 41555 by Raj Singh last updated on 09/Aug/18 Answered by alex041103 last updated on 09/Aug/18 $$\mathrm{1}.\:\int\frac{{ln}\:{x}}{\left(\mathrm{1}+{ln}\:{x}\right)^{\mathrm{2}} }\:{dx} \\ $$$${let}\:{x}={e}^{{u}} \:{dx}={e}^{{u}} {du}. \\ $$$$\int\frac{\mathrm{1}+{u}−\mathrm{1}}{\left(\mathrm{1}+{u}\right)^{\mathrm{2}}…
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Question Number 41518 by maxmathsup by imad last updated on 08/Aug/18 $${calculate}\:\:{A}_{{n}} =\:\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{1}−{t}^{\mathrm{2}} \right)^{{n}} {dt}\:\:\:{with}\:{n}\:{integr}\:{natural} \\ $$ Answered by alex041103 last updated on…
Question Number 41516 by maxmathsup by imad last updated on 08/Aug/18 $${calculate}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\frac{{ln}\left(\mathrm{1}+{x}\right)}{\left(\mathrm{1}+{x}\right)^{\mathrm{4}} }\:{dx} \\ $$ Commented by turbo msup by abdo last updated…
Question Number 41515 by maxmathsup by imad last updated on 08/Aug/18 $$\left.{l}\left.{et}\:\:{f}_{{n}} \left({x}\right)\:=\frac{{sin}\left(\mathrm{2}\left({n}+\mathrm{1}\right){x}\right)}{{sinx}}\:{if}\:\:{x}\in\right]\mathrm{0},\frac{\pi}{\mathrm{2}}\right]\:{and}\:{f}_{{n}} \left(\mathrm{0}\right)=\mathrm{2}\left({n}+\mathrm{1}\right)\:\:{let} \\ $$$${u}_{{n}} =\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:{f}_{{n}} \left({x}\right){dx} \\ $$$$\left.\mathrm{1}\right)\:{prove}\:{that}\:\:\forall{n}\:{fromN}\:\:{u}_{{n}+\mathrm{1}} −{u}_{{n}} =\mathrm{2}\frac{\left(−\mathrm{1}\right)^{{n}+\mathrm{1}} }{\mathrm{2}{n}+\mathrm{3}}…
Question Number 41514 by maxmathsup by imad last updated on 08/Aug/18 $${find}\:\:\:\int\:\:{cos}\left({lnx}\right){dx}\: \\ $$ Answered by alex041103 last updated on 09/Aug/18 $${We}\:{know}\:{that}\:{cos}\left(\theta\right)={Re}\left({e}^{{i}\theta} \right) \\ $$$${We}\:{will}\:{assume}\:{x}\epsilon\mathbb{R}.…
Question Number 107051 by Ar Brandon last updated on 08/Aug/20 $$\int_{\mathrm{0}} ^{\pi} \frac{\mathrm{cosx}}{\mathrm{2}+\mathrm{sin}^{\mathrm{2}} \mathrm{x}}\mathrm{dx} \\ $$ Answered by Dwaipayan Shikari last updated on 08/Aug/20 $$\int_{\mathrm{0}}…