Question Number 20293 by tammi last updated on 25/Aug/17 $$\int\sqrt{\frac{{a}+{x}}{{x}}{dx}} \\ $$ Answered by $@ty@m last updated on 25/Aug/17 $$=\int\frac{{a}+{x}}{\:\sqrt{{x}\left({a}+{x}\right)}}{dx} \\ $$$$=\int\frac{{a}+{x}}{\:\sqrt{{ax}+{x}^{\mathrm{2}} }}{dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\mathrm{2}{x}+{a}+{a}}{\:\sqrt{{ax}+{x}^{\mathrm{2}}…
Question Number 85828 by M±th+et£s last updated on 25/Mar/20 $$\int\frac{\mathrm{1}}{{x}+{cot}\left({x}\right)}\:{dx} \\ $$ Answered by mind is power last updated on 26/Mar/20 $${not}\:{found}\:\:{only}\:{withe}\:{Series}\: \\ $$$${have}\:{you}\:{a}\:{solutions}\:{sir}\:? \\…
Question Number 20292 by tammi last updated on 25/Aug/17 $$\int\frac{{dx}}{\left({x}+\mathrm{1}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} +\sqrt{{x}−\mathrm{1}}} \\ $$ Answered by $@ty@m last updated on 25/Aug/17 $$=\int\frac{{dx}}{\:\sqrt{{x}+\mathrm{1}}+\sqrt{{x}−\mathrm{1}}} \\ $$$$=\int\frac{\sqrt{{x}+\mathrm{1}}−\sqrt{{x}−\mathrm{1}}}{\left({x}+\mathrm{1}\right)−\left({x}−\mathrm{1}\right)}{dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int\sqrt{{x}+\mathrm{1}}{dx}−\frac{\mathrm{1}}{\mathrm{2}}\int\sqrt{{x}−\mathrm{1}}{dx}…
Question Number 20291 by tammi last updated on 25/Aug/17 $$\sqrt{\frac{\mathrm{1}−{x}}{\mathrm{1}+{x}}\:{dx}} \\ $$ Answered by $@ty@m last updated on 25/Aug/17 $$=\int\frac{\mathrm{1}−{x}}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}{dx} \\ $$$$=\int\frac{{dx}}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}+\frac{\mathrm{1}}{\mathrm{2}}\int\frac{−\mathrm{2}{x}}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}{dx}…
Question Number 151360 by talminator2856791 last updated on 20/Aug/21 $$\: \\ $$$$\:\:\mathrm{find}\:\mathrm{max}\left(\Re\left(\mathrm{I}\right)+\Im\left(\mathrm{I}\right)\right)\:\mathrm{for}\:\mathrm{the}\:\mathrm{integral} \\ $$$$\:\:\mathrm{I}\:=\:\int_{{z}} ^{\:{z}+\mathrm{1}} \:\mathrm{cos}\left(\mathrm{cos}\left(\mathrm{cos}\left({x}^{\mathrm{cos}\left(\mathrm{cos}\left(\mathrm{cos}\left({x}\right)\right)\right)} \right)\right)\right){dx} \\ $$$$\:\:{z}\:\in\:\mathbb{R} \\ $$$$\: \\ $$ Terms of…
Question Number 20281 by ajfour last updated on 25/Aug/17 $${Compute}\:{the}\:{volume}\:{bounded}\:{by} \\ $$$${the}\:{surfaces}:\:{y}={x}^{\mathrm{2}} ,\:{x}={y}^{\mathrm{2}} ,\:{z}=\mathrm{0}, \\ $$$${z}=\mathrm{12}+{y}−{x}^{\mathrm{2}} .\:\:\:\:\:\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left[\:{Ans}.\:\:\:\:\frac{\mathrm{549}}{\mathrm{144}}\right] \\ $$ Terms of Service Privacy…
Question Number 151345 by peter frank last updated on 20/Aug/21 $$\int_{\mathrm{0}} ^{\pi} \frac{\mathrm{e}^{\mathrm{cos}\:\mathrm{x}} }{\mathrm{e}^{\mathrm{cos}\:\mathrm{x}} +\mathrm{e}^{−\mathrm{cos}\:\mathrm{x}} }\mathrm{dx} \\ $$ Answered by Olaf_Thorendsen last updated on 20/Aug/21…
Question Number 85813 by jagoll last updated on 25/Mar/20 $$\int\:\mathrm{x}^{\mathrm{2}} \:\sqrt{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }\:\mathrm{dx}\:? \\ $$ Commented by jagoll last updated on 25/Mar/20 $$\int\:\frac{\mathrm{1}}{\mathrm{2}}\mathrm{x}\:\left(\mathrm{2x}\sqrt{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }\:\right)\:\mathrm{dx}\:=\: \\ $$$$\int\:\frac{\mathrm{1}}{\mathrm{2}}\mathrm{x}\:\sqrt{\mathrm{1}+\mathrm{x}^{\mathrm{2}}…
Question Number 151341 by peter frank last updated on 20/Aug/21 $$\int_{\alpha} ^{\beta} \frac{\mathrm{dx}}{\mathrm{x}\sqrt{\left(\mathrm{x}−\alpha\right)\left(\beta−\mathrm{x}\right)\:}}=\frac{\pi}{\:\sqrt{\alpha\beta}} \\ $$$$\mathrm{where}\:\alpha,\beta\:>\mathrm{0} \\ $$ Answered by Kamel last updated on 20/Aug/21 $$…
Question Number 85807 by jagoll last updated on 25/Mar/20 $$\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\:\frac{\mathrm{x}^{\mathrm{2}} \:\mathrm{dx}}{\:\sqrt{\mathrm{1}−\mathrm{x}^{\mathrm{4}} }} \\ $$ Answered by Joel578 last updated on 25/Mar/20 $${I}\:\:=\:\int_{\mathrm{0}} ^{\:\mathrm{1}}…