Question Number 73155 by MJS last updated on 06/Nov/19 $$\mathrm{reposting}\:\mathrm{a}\:\mathrm{former}\:\mathrm{question}… \\ $$$$\int\frac{\sqrt[{\mathrm{5}}]{{x}}−\mathrm{1}}{\:\sqrt{{x}}+\mathrm{1}}{dx}= \\ $$$$\:\:\:\:\:\left[{t}=\sqrt[{\mathrm{10}}]{{x}}\:\rightarrow\:{dx}=\mathrm{10}\sqrt[{\mathrm{10}}]{{x}^{\mathrm{9}} }{dx}\right] \\ $$$$=\mathrm{10}\int\frac{{t}^{\mathrm{9}} \left({t}−\mathrm{1}\right)}{{t}^{\mathrm{4}} −{t}^{\mathrm{3}} +{t}^{\mathrm{2}} −{t}+\mathrm{1}}{dt}= \\ $$$$=\mathrm{10}\int\left({t}^{\mathrm{6}} −{t}^{\mathrm{4}} −{t}\right){dt}+\mathrm{10}\int\frac{{t}\left({t}^{\mathrm{2}}…
Question Number 138691 by liberty last updated on 16/Apr/21 $$\int\:\mathrm{cos}\:\mathrm{2}{x}\:\sqrt{\mathrm{1}+\mathrm{sin}\:^{\mathrm{2}} {x}}\:{dx}\:=? \\ $$ Answered by mathmax by abdo last updated on 17/Apr/21 $$\mathrm{I}=\int\:\mathrm{cos}\left(\mathrm{2x}\right)\sqrt{\mathrm{1}+\mathrm{sin}^{\mathrm{2}} \mathrm{x}}\mathrm{dx}\:\Rightarrow\mathrm{I}=\int\:\mathrm{cos}\left(\mathrm{2x}\right)\sqrt{\mathrm{1}+\frac{\mathrm{1}−\mathrm{cos}\left(\mathrm{2x}\right)}{\mathrm{2}}}\mathrm{dx} \\…
Question Number 138690 by KwesiDerek last updated on 16/Apr/21 $$\left(\frac{\mathrm{x}}{\mathrm{12}}\right)^{\mathrm{log}_{\sqrt{\mathrm{3}}} \mathrm{x}} =\left(\frac{\mathrm{x}}{\mathrm{18}}\right)^{\mathrm{log}_{\sqrt{\mathrm{2}}} \mathrm{x}} \\ $$$$\boldsymbol{\mathrm{find}}\:\boldsymbol{\mathrm{x}} \\ $$ Commented by liberty last updated on 16/Apr/21 $${x}=\mathrm{36}\:\Rightarrow\left(\frac{\mathrm{36}}{\mathrm{12}}\right)^{\mathrm{log}\:_{\sqrt{\mathrm{3}}}…
Question Number 7612 by Tawakalitu. last updated on 06/Sep/16 $${If}\:\:{ax}\:+\:{by}\:+\:{cz}\:=\:\mathrm{0},\: \\ $$$${and},\:\:{a}^{\mathrm{2}} {x}\:+\:{b}^{\mathrm{2}} {y}\:+\:{c}^{\mathrm{2}} {z}\:=\:\mathrm{0} \\ $$$${Find}\:{the}\:{ratio}\:\:{x}:{y}:{z} \\ $$ Commented by Rasheed Soomro last updated…
Question Number 7611 by Tawakalitu. last updated on 06/Sep/16 $${Find}\:{the}\:{square}\:{root}\:{of}\: \\ $$$$\mathrm{121}{x}^{\mathrm{6}} \:+\:\mathrm{44}{x}^{\mathrm{5}} \:−\:\mathrm{18}{x}^{\mathrm{4}} \:+\:\mathrm{18}{x}^{\mathrm{3}} \:+\:\mathrm{5}{x}^{\mathrm{2}} \:−\:\mathrm{2}{x}\:+\:\mathrm{1} \\ $$ Answered by Rasheed Soomro last updated…
Question Number 73147 by amaramariche last updated on 06/Nov/19 $$\int\frac{{x}−\mathrm{6}}{{x}^{\mathrm{3}} +\mathrm{1}}{dx} \\ $$$$ \\ $$ Commented by mathmax by abdo last updated on 07/Nov/19 $${let}\:{A}\:=\int\:\frac{{x}−\mathrm{6}}{{x}^{\mathrm{3}}…
Question Number 7610 by Tawakalitu. last updated on 06/Sep/16 $${obtain}\:{the}\:{value}\:{of}\: \\ $$$$\left({a}^{\mathrm{1}/\mathrm{2}} +{b}^{\mathrm{1}/\mathrm{2}} +{c}^{\mathrm{1}/\mathrm{2}} \right)\left({a}^{\mathrm{1}/\mathrm{2}} −{b}^{\mathrm{1}/\mathrm{2}} +{c}^{\mathrm{1}/\mathrm{2}} \right)\left({a}^{\mathrm{1}/\mathrm{2}} −{b}^{\mathrm{1}/\mathrm{2}} +{c}^{\mathrm{1}/\mathrm{2}} \right)\left({b}^{\mathrm{1}/\mathrm{2}} +{c}^{\mathrm{1}/\mathrm{2}} −{a}^{\mathrm{1}/\mathrm{2}} \right)\: \\…
Question Number 138683 by mnjuly1970 last updated on 16/Apr/21 $$\:\:\:\:\:\:\:\:\:\:\:\:\:…{nice}\:..\:…\:…\:{calculus}… \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:{find}\:{the}\:{value}\:{of}: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\Theta=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} {sin}^{\mathrm{2}} \left({n}\right)}{{n}}=? \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:……………………. \\ $$ Answered by Dwaipayan…
Question Number 73144 by mathmax by abdo last updated on 06/Nov/19 $${calculte}\:\int\:\:\frac{{x}+\sqrt{\mathrm{2}+{x}^{\mathrm{2}} }}{{x}+\mathrm{1}−\sqrt{\mathrm{2}+{x}^{\mathrm{2}} }}{dx} \\ $$ Commented by mathmax by abdo last updated on 07/Nov/19…
Question Number 7609 by Tawakalitu. last updated on 06/Sep/16 $${Divide}\: \\ $$$${a}^{\mathrm{5}/\mathrm{2}} \:−\:\mathrm{5}{a}^{\mathrm{2}} {b}^{\mathrm{1}/\mathrm{3}} \:+\:\mathrm{10}{a}^{\mathrm{3}/\mathrm{2}} \mathrm{6}^{\mathrm{2}/\mathrm{3}} \:−\:\mathrm{10}{ab}\:+\:\mathrm{5}{a}^{\mathrm{1}/\mathrm{2}} {b}^{\mathrm{4}/\mathrm{3}} \:−\:{b}^{\mathrm{5}/\mathrm{3}} \\ $$$${by}\:\:{a}^{\mathrm{1}/\mathrm{2}} \:−\:{b}^{\mathrm{1}/\mathrm{3}} \\ $$ Answered…