Question Number 137275 by mnjuly1970 last updated on 31/Mar/21 $$\:\:\:\:\:\:\:\:\:\:\:\:\:……{complex}\:\:{analysis}….. \\ $$$$\:\:\:\:{if}\:,\:\:{f}\left(\alpha,{n},{x}\right)=\frac{{d}^{\:{n}} }{{dx}^{{n}} }\left(\alpha^{{x}} \right)\:\:,\:{x}\in\mathbb{C} \\ $$$$\:\:\:\:\:\alpha\in\mathbb{C}−\left\{\mathrm{0}\right\}\:,\:{n}\in\mathbb{C}−\mathbb{Z}^{−} \cup\left\{\mathrm{0}\right\} \\ $$$$\:\:\:\:\:{and}\:\:{g}\left({n},{x}\right)=\int_{\mathrm{0}} ^{\:\mathrm{1}} {f}\left(\alpha,{n},{x}\right){d}\alpha \\ $$$$\:\:\:\:\:{then}\:\:{find}\:\:{the}\:{value}\:{of}\:… \\…
Question Number 6180 by FilupSmith last updated on 17/Jun/16 $$\frac{{df}\left({x}\right)}{{dx}}={f}\left({x}\right)+{x} \\ $$$$\mathrm{Solve}\:{f}\left({x}\right) \\ $$ Commented by 123456 last updated on 17/Jun/16 $$\frac{{df}}{{dx}}−{f}={x} \\ $$$${f}={f}_{{n}} +{f}_{{g}}…
Question Number 137249 by bemath last updated on 31/Mar/21 $$\int\:\frac{\left(\mathrm{3sin}\:\mathrm{x}+\mathrm{2}\right)}{\left(\mathrm{2sin}\:\mathrm{x}+\mathrm{3}\right)^{\mathrm{2}} }\:\mathrm{dx}\:=? \\ $$ Answered by bemath last updated on 31/Mar/21 $$\mathrm{let}\:\mathrm{tan}\:\left(\frac{\mathrm{x}}{\mathrm{2}}\right)=\mathrm{u}\: \begin{cases}{\mathrm{dx}=\frac{\mathrm{2}}{\mathrm{u}^{\mathrm{2}} +\mathrm{1}}\:\mathrm{du}}\\{\mathrm{sin}\:\mathrm{x}=\frac{\mathrm{2u}}{\mathrm{u}^{\mathrm{2}} +\mathrm{1}}}\end{cases} \\…
Question Number 137251 by mnjuly1970 last updated on 31/Mar/21 $$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:……\mathscr{A}{dvanced}\:\:…\:\:{calculus}…… \\ $$$$\:\:\:\:\boldsymbol{\phi}=\int_{\mathrm{0}} ^{\:\mathrm{1}} {x}^{\mathrm{2}} {ln}\left({x}\right){ln}\left(\mathrm{1}−{x}\right){dx}=??? \\ $$$$ \\ $$ Answered by Ar Brandon last updated…
Question Number 6152 by sanusihammed last updated on 16/Jun/16 $${Differentiate}\:\:\:{s}\:=\:{ut}\:+\:\frac{\mathrm{1}}{\mathrm{2}}{at}^{\mathrm{2}} \:\:\:\:\:\:{from}\:{the}\:{first}\:{principle} \\ $$$$ \\ $$ Answered by Rasheed Soomro last updated on 16/Jun/16 $${s}\:=\:{ut}\:+\:\frac{\mathrm{1}}{\mathrm{2}}{at}^{\mathrm{2}} \:\:\:…
Question Number 137226 by SLVR last updated on 31/Mar/21 Answered by MJS_new last updated on 31/Mar/21 $${f}\left({x}\right)=\mathrm{2ln}\:{x} \\ $$$${x}^{\mathrm{3}} −\mathrm{6}{x}^{\mathrm{2}} +\mathrm{11}{x}−\mathrm{6}=\left({x}−\mathrm{1}\right)\left({x}−\mathrm{2}\right)\left({x}−\mathrm{3}\right) \\ $$$$\mathrm{the}\:\mathrm{curves}\:\mathrm{intersect}\:\mathrm{at}\:{x}=\mathrm{1} \\ $$$$\underset{\mathrm{0}}…
Question Number 137190 by mnjuly1970 last updated on 30/Mar/21 $$\:\:\:\:\:\:\:\:\:\:\:\:\:\:….{advanced}\:…….\:\:{calculus}….. \\ $$$$\:\:\:\:\:\:\:\:{prove}\:{that}::: \\ $$$$\boldsymbol{\phi}=\int_{\mathrm{0}} ^{\:\mathrm{1}} {ln}\left({ln}\left(\frac{\mathrm{1}}{{x}}\right)\right).\frac{{dx}}{\:\sqrt{{ln}\left(\frac{\mathrm{1}}{{x}}\right)}}\:=−\sqrt{\pi}\:\left(\gamma+{ln}\left(\mathrm{4}\right)\right) \\ $$$$ \\ $$ Answered by Dwaipayan Shikari last…
Question Number 137129 by Chhing last updated on 30/Mar/21 $$ \\ $$$$\:\left(\mathrm{x}^{\mathrm{2}} −\mathrm{1}\right)\mathrm{y}''−\mathrm{n}\left(\mathrm{n}+\mathrm{1}\right)\mathrm{y}=\mathrm{0}\:\:\:,\:\:\mathrm{n}\in\mathbb{N} \\ $$$$\mathrm{Find}\:\mathrm{solution}\:\mathrm{that}\:\mathrm{can}\:\mathrm{be}\:\mathrm{expanded}\:\mathrm{in}\:\mathrm{series} \\ $$$$\mathrm{help}\:\mathrm{me} \\ $$ Answered by mathmax by abdo last…
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Question Number 71550 by mhmd last updated on 17/Oct/19 Answered by MJS last updated on 17/Oct/19 $$\mathrm{it}'\:\mathrm{wrong},\:\mathrm{it}\:\mathrm{must}\:\mathrm{be} \\ $$$$\begin{pmatrix}{{n}+\mathrm{1}}\\{{k}}\end{pmatrix}\:=\begin{pmatrix}{{n}}\\{{k}}\end{pmatrix}\:+\begin{pmatrix}{{n}}\\{{k}−\mathrm{1}}\end{pmatrix} \\ $$$$\begin{pmatrix}{{n}}\\{{k}}\end{pmatrix}\:+\begin{pmatrix}{{n}}\\{{k}−\mathrm{1}}\end{pmatrix}\:=\frac{{n}!}{{k}!\left({n}−{k}\right)!}+\frac{{n}!}{\left({k}−\mathrm{1}\right)!\left({n}−{k}+\mathrm{1}\right)!}= \\ $$$$=\frac{{n}!\left({n}−{k}+\mathrm{1}\right)}{\left({k}−\mathrm{1}\right)!\left({n}−{k}+\mathrm{1}\right)!{k}}+\frac{{n}!{k}}{\left({k}−\mathrm{1}\right)!\left({n}−{k}+\mathrm{1}\right)!{k}}= \\ $$$$=\frac{{n}!\left({n}−{k}+\mathrm{1}\right)+{n}!{k}}{\left({k}−\mathrm{1}\right)!\left({n}−{k}+\mathrm{1}\right)!{k}}=\frac{{n}!\left({n}+\mathrm{1}\right)}{{k}!\left({n}+\mathrm{1}−{k}\right)!}=…