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Author: Tinku Tara

4-u-u-2-5u-6-du-

Question Number 85362 by sahnaz last updated on 21/Mar/20 $$\int\frac{−\mathrm{4}−\mathrm{u}}{\mathrm{u}^{\mathrm{2}} +\mathrm{5u}+\mathrm{6}}\mathrm{du} \\ $$ Answered by john santu last updated on 22/Mar/20 $$\int\:\frac{{du}}{{u}+\mathrm{3}}\:−\int\:\frac{\mathrm{2}}{{u}+\mathrm{2}}\:{du}\:=\: \\ $$$$\:\mathrm{ln}\:\mid{u}+\mathrm{3}\mid\:−\:\mathrm{2}\:\mathrm{ln}\:\mid{u}+\mathrm{2}\mid\:+\:{c}\: \\…

x-1-3-x-12-

Question Number 85358 by jagoll last updated on 21/Mar/20 $$\sqrt[{\mathrm{3}\:\:}]{\mathrm{x}\:}\:+\:\sqrt{\mathrm{x}}\:=\:\mathrm{12} \\ $$ Answered by john santu last updated on 21/Mar/20 $${x}\:=\:{t}^{\mathrm{6}} \\ $$$$\Rightarrow\:{t}^{\mathrm{2}} +{t}^{\mathrm{3}} −\mathrm{12}=\mathrm{0}…

Question-19822

Question Number 19822 by virus last updated on 16/Aug/17 Answered by ajfour last updated on 16/Aug/17 $$\left[\mathrm{xsin}\:\theta_{\mathrm{1}} −\left({l}−\mathrm{x}\right)\mathrm{sin}\:\theta_{\mathrm{2}} \right]\mathrm{g}={l}\mathrm{a} \\ $$$$\:\mathrm{a}=\left[\frac{\mathrm{x}}{{l}}\mathrm{sin}\:\theta_{\mathrm{1}} −\frac{\left({l}−\mathrm{x}\right)}{{l}}\mathrm{sin}\:\theta_{\mathrm{2}} \right]\mathrm{g}\:\:. \\ $$…

Question-150889

Question Number 150889 by mathdanisur last updated on 16/Aug/21 Answered by dumitrel last updated on 16/Aug/21 $$\Leftrightarrow{a}^{\mathrm{2}} +{a}\left({b}+{c}+{d}\right)+\left({b}+{c}+{d}\right)^{\mathrm{2}} \geqslant\mathrm{0}\Leftrightarrow \\ $$$$\left(\frac{{a}}{\mathrm{2}}+{b}+{c}+{d}\right)^{\mathrm{2}} +\frac{\mathrm{3}{a}^{\mathrm{2}} }{\mathrm{4}}\geqslant\mathrm{0} \\ $$$$,,=''\:{for}\:{a}=\mathrm{0};\:{b}+{c}+{d}=\mathrm{0}…

Question-85355

Question Number 85355 by TawaTawa1 last updated on 21/Mar/20 Commented by mathmax by abdo last updated on 21/Mar/20 $${I}\:=\int\:{arctan}\left(\sqrt{\mathrm{1}+\sqrt{{x}}}\right){dx}\:\:{changement}\:\sqrt{\mathrm{1}+\sqrt{{x}}}={t}\:{give} \\ $$$$\mathrm{1}+\sqrt{{x}}={t}^{\mathrm{2}} \:\Rightarrow\sqrt{{x}}={t}^{\mathrm{2}} −\mathrm{1}\:\Rightarrow{x}\:=\left({t}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} \:\Rightarrow{dx}\:=\mathrm{2}\left(\mathrm{2}{t}\right)\left({t}^{\mathrm{2}}…