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proporsed-by-m-njuly-1970-0-1-ln-x-1-1-3-x-1-x-2-dx-solution-let-I-1-3-0-1-ln-x-1-x-1-x-2-dx-I-1-3-0-1-ln-1-x-x-1-dx-0-1-ln-x-1-x-2-dx-A-B-let-A-1-3

Question Number 112203 by mathdave last updated on 06/Sep/20 $${proporsed}\:{by}\:{m}.{njuly}\:\mathrm{1970} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{ln}\left(\sqrt[{\mathrm{3}}]{{x}+\mathrm{1}}\right)}{\left({x}+\mathrm{1}\right)\left({x}+\mathrm{2}\right)}{dx} \\ $$$${solution} \\ $$$${let}\:{I}=\frac{\mathrm{1}}{\mathrm{3}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{ln}\left({x}+\mathrm{1}\right)}{\left({x}+\mathrm{1}\right)\left({x}+\mathrm{2}\right)}{dx} \\ $$$${I}=\frac{\mathrm{1}}{\mathrm{3}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{ln}\left(\mathrm{1}+{x}\right)}{{x}+\mathrm{1}}{dx}−\int_{\mathrm{0}} ^{\mathrm{1}}…

2-1-2-y-5-1-2-y-2-3-could-you-help-me-

Question Number 46652 by gunay last updated on 29/Oct/18 $$\mathrm{2}\frac{\mathrm{1}}{\mathrm{2}}\mathrm{y}+\mathrm{5}\frac{\mathrm{1}}{\mathrm{2}}\mathrm{y}−\mathrm{2}=\mathrm{3} \\ $$$$\mathrm{could}\:\mathrm{you}\:\mathrm{help}\:\mathrm{me} \\ $$ Answered by tanmay.chaudhury50@gmail.com last updated on 29/Oct/18 $$\left(\mathrm{2}+\frac{\mathrm{1}}{\mathrm{2}}\right){y}+\left(\mathrm{5}+\frac{\mathrm{1}}{\mathrm{2}}\right){y}=\mathrm{5} \\ $$$${y}\left(\mathrm{2}+\frac{\mathrm{1}}{\mathrm{2}}+\mathrm{5}+\frac{\mathrm{1}}{\mathrm{2}}\right)=\mathrm{5} \\…

Solve-in-C-x-y-z-1-x-1-y-1-z-

Question Number 177702 by lapache last updated on 08/Oct/22 $${Solve}\:{in}\:\mathbb{C} \\ $$$$\begin{cases}{{x}+{y}={z}}\\{\frac{\mathrm{1}}{{x}}+\frac{\mathrm{1}}{{y}}=\frac{\mathrm{1}}{{z}}}\end{cases} \\ $$ Answered by Frix last updated on 08/Oct/22 $$\left(\mathrm{1}\right)\:{x}={z}−{y} \\ $$$$\left(\mathrm{2}\right)\:{x}=\frac{{yz}}{{y}−{z}} \\…

Question-177665

Question Number 177665 by Ahmed777hamouda last updated on 07/Oct/22 Answered by Ar Brandon last updated on 07/Oct/22 $$\int_{−\infty} ^{\infty} \mathrm{sin}{x}^{\mathrm{2}} {dx}=\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{sin}{t}}{\:\sqrt{{t}}}{dt}=\frac{\pi}{\mathrm{2}\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}\right)\mathrm{sin}\left(\frac{\pi}{\mathrm{2}}\centerdot\frac{\mathrm{1}}{\mathrm{2}}\right)}=\sqrt{\frac{\pi}{\mathrm{2}}} \\ $$…

Question-112133

Question Number 112133 by shwebotetun last updated on 06/Sep/20 Answered by som(math1967) last updated on 06/Sep/20 $$\left(−\mathrm{2b}\right)^{\mathrm{2}} +\mathrm{20b}+\mathrm{14}=\left(−\mathrm{2c}\right)^{\mathrm{2}} +\mathrm{20c}+\mathrm{14} \\ $$$$\mathrm{4b}^{\mathrm{2}} −\mathrm{4c}^{\mathrm{2}} +\mathrm{20}\left(\mathrm{b}−\mathrm{c}\right)=\mathrm{0} \\ $$$$\mathrm{4}\left\{\left(\mathrm{b}−\mathrm{c}\right)\left(\mathrm{b}+\mathrm{c}\right)+\mathrm{5}\left(\mathrm{b}−\mathrm{c}\right)\right\}=\mathrm{0}…