Question Number 6027 by FilupSmith last updated on 10/Jun/16 $${x}^{{x}} ={e}^{{x}\mathrm{ln}\:{x}} \\ $$$${e}^{{x}} =\mathrm{1}+{x}+\frac{{x}^{\mathrm{2}} }{\mathrm{2}!}+\frac{{x}^{\mathrm{3}} }{\mathrm{3}!}+… \\ $$$$\therefore\:{e}^{{x}\mathrm{ln}\:{x}} \:=\:\mathrm{1}+{x}\mathrm{ln}\left({x}\right)+\frac{\left({x}\mathrm{ln}\:{x}\right)^{\mathrm{2}} }{\mathrm{2}!}+… \\ $$$${e}^{{x}\mathrm{ln}\:{x}} \:=\:\frac{\left({x}\mathrm{ln}\:{x}\right)^{\mathrm{0}} }{\mathrm{0}!}+\frac{\left({x}\mathrm{ln}\:{x}\right)^{\mathrm{1}} }{\mathrm{1}!}+\frac{\left({x}\mathrm{ln}\:{x}\right)^{\mathrm{2}}…
Question Number 137093 by mnjuly1970 last updated on 29/Mar/21 $$\:\:\:\:\:\:\:\:\:…..{advanced}\:\:\:\:{calculus}…. \\ $$$$\:\:\:\:\:{please}\:\:{evaluate}::\:\: \\ $$$$\:\:\:\:\:\mathrm{1}:\:\underset{{n}=\mathrm{1}\:} {\overset{\infty} {\sum}}\frac{{H}_{{n}} }{\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{2}} }\:=? \\ $$$$\:\:\:\:\mathrm{2}:\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{{xln}\left({x}\right){ln}\left(\mathrm{1}−{x}\right)}{\mathrm{1}−{x}}{dx}=? \\ $$$$\:\:{note}:{H}_{{n}} =\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{3}}+…+\frac{\mathrm{1}}{{n}}=\int_{\mathrm{0}}…
Question Number 6016 by sanusihammed last updated on 09/Jun/16 $$\int\left(\frac{\mathrm{2}{x}\:+\:\mathrm{5}}{\:\sqrt{\mathrm{3}\:+\:\mathrm{4}{x}\:−\:\mathrm{5}{x}^{\mathrm{2}} }}\right){dx} \\ $$$$ \\ $$$${please}\:{help}. \\ $$ Answered by Yozzii last updated on 09/Jun/16 $${Let}\:{u}=\mathrm{3}+\mathrm{4}{x}−\mathrm{5}{x}^{\mathrm{2}}…
Question Number 137083 by mnjuly1970 last updated on 29/Mar/21 $$\:\:\:\:\:\boldsymbol{\phi}=\int^{\:} {sin}\left(\frac{\mathrm{2}}{{x}}\right)\sqrt{\mathrm{1}+{cos}^{\mathrm{2}} \left(\frac{\mathrm{1}}{{x}}\right)}\:\frac{{dx}}{{x}^{\mathrm{2}} } \\ $$ Answered by Ar Brandon last updated on 29/Mar/21 $$\emptyset=\int\mathrm{sin}\left(\frac{\mathrm{2}}{\mathrm{x}}\right)\sqrt{\mathrm{1}+\mathrm{cos}^{\mathrm{2}} \left(\frac{\mathrm{1}}{\mathrm{x}}\right)}\frac{\mathrm{dx}}{\mathrm{x}^{\mathrm{2}}…
Question Number 137055 by mathmax by abdo last updated on 29/Mar/21 $$\mathrm{find}\:\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\frac{\mathrm{dx}}{\left(\mathrm{1}+\mathrm{cosx}\:+\mathrm{3sinx}\right)^{\mathrm{2}} } \\ $$ Commented by MJS_new last updated on 30/Mar/21 $$\mathrm{0}\leqslant\left(\mathrm{1}+\mathrm{cos}\:{x}\:+\mathrm{3sin}\:{x}\right)^{\mathrm{2}}…
Question Number 137048 by mnjuly1970 last updated on 29/Mar/21 $$\:\:\:\:\:\:\:\:\:\:\:……..{advanced}\:\:…\:…\:…\:{calculus}……. \\ $$$$\:\:\:\:{evaluate}:::: \\ $$$$\:\:\:\:\:\:\boldsymbol{\chi}=\int_{\mathrm{0}} ^{\:\infty} {x}^{\mathrm{2}} \:{e}^{−{x}^{\mathrm{2}} } {ln}\left({x}\right)=??? \\ $$$$\:\: \\ $$ Answered by…
Question Number 71499 by petrochengula last updated on 16/Oct/19 Commented by petrochengula last updated on 16/Oct/19 $$\mathrm{help}\:\mathrm{2a} \\ $$ Answered by MJS last updated on…
Question Number 71490 by oyemi kemewari last updated on 16/Oct/19 Commented by mathmax by abdo last updated on 16/Oct/19 $${let}\:{A}=\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \:\frac{\mathrm{1}+\sqrt{\mathrm{3}}}{\left(^{\mathrm{4}} \sqrt{\left(\mathrm{1}+{x}\right)^{\mathrm{2}} \left(\mathrm{1}−{x}\right)^{\mathrm{6}} }\right.}{dx}\:\Rightarrow{A}=\left(\mathrm{1}+\sqrt{\mathrm{3}}\right)\int_{\mathrm{0}}…
Question Number 137004 by Mathspace last updated on 28/Mar/21 $${find}\:\int\:\frac{\sqrt{{x}}}{\:\sqrt{{x}−\mathrm{1}}+\sqrt{{x}+\mathrm{1}}}{dx} \\ $$ Answered by aleks041103 last updated on 28/Mar/21 $$\frac{\sqrt{{x}}}{\:\sqrt{{x}−\mathrm{1}}+\sqrt{{x}+\mathrm{1}}}= \\ $$$$=\frac{\sqrt{{x}}}{\:\sqrt{{x}−\mathrm{1}}+\sqrt{{x}+\mathrm{1}}}\:\frac{\sqrt{{x}+\mathrm{1}}−\sqrt{{x}−\mathrm{1}}}{\:\sqrt{{x}+\mathrm{1}}−\sqrt{{x}−\mathrm{1}}}= \\ $$$$=\sqrt{{x}}\left(\sqrt{{x}+\mathrm{1}}−\sqrt{{x}−\mathrm{1}}\right) \\…
Question Number 137003 by Mathspace last updated on 28/Mar/21 $${find}\:\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:\:\frac{{cos}^{\mathrm{5}} {t}}{{cos}\left(\mathrm{5}{t}\right)}{dt} \\ $$ Answered by bobhans last updated on 29/Mar/21 $$\mathrm{cos}\:\left(\mathrm{5t}\right)=\mathrm{cos}\:^{\mathrm{5}} \mathrm{t}−\mathrm{10sin}\:^{\mathrm{2}} \mathrm{t}\:\mathrm{cos}\:^{\mathrm{3}}…