Question Number 85601 by sahnaz last updated on 23/Mar/20 $$\int\frac{\mathrm{4u}}{\mathrm{4u}^{\mathrm{2}} −\mathrm{4u}+\mathrm{1}}\mathrm{du} \\ $$ Commented by Tony Lin last updated on 23/Mar/20 $$\int\frac{\mathrm{4}{u}}{\mathrm{4}{u}^{\mathrm{2}} −\mathrm{4}{u}+\mathrm{1}}{du} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\mathrm{8}{u}−\mathrm{4}}{\mathrm{4}{u}^{\mathrm{2}}…
Question Number 85592 by sahnaz last updated on 23/Mar/20 $$\int\frac{\left(\mathrm{u}+\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{u}^{\mathrm{3}} +\mathrm{u}}\mathrm{du} \\ $$ Answered by john santu last updated on 23/Mar/20 $$\int\:\frac{\mathrm{1}}{{x}}{dx}\:+\:\int\:\frac{\mathrm{2}}{{x}^{\mathrm{2}} +\mathrm{1}}\:{dx} \\…
Question Number 85596 by M±th+et£s last updated on 23/Mar/20 $$\int\frac{\sqrt{{x}+\mathrm{1}}−\mathrm{1}}{\:\sqrt{{x}−\mathrm{1}}+\mathrm{1}}\:{dx} \\ $$ Commented by mathmax by abdo last updated on 23/Mar/20 $${A}\:=\int\:\:\frac{\sqrt{{x}+\mathrm{1}}−\mathrm{1}}{\:\sqrt{{x}−\mathrm{1}}+\mathrm{1}}{dx}\:\:\:\:{chagement}\:\sqrt{{x}−\mathrm{1}}+\mathrm{1}\:={t}\:{give}\:\sqrt{{x}−\mathrm{1}}={t}−\mathrm{1}\:\Rightarrow \\ $$$${x}−\mathrm{1}\:=\left({t}−\mathrm{1}\right)^{\mathrm{2}} \:\Rightarrow{dx}\:=\mathrm{2}\left({t}−\mathrm{1}\right){dt}\:\Rightarrow…
Question Number 85591 by sahnaz last updated on 23/Mar/20 $$\int\frac{\mathrm{1}+\mathrm{4u}}{−\mathrm{4u}^{\mathrm{2}} +\mathrm{2u}+\mathrm{2}}\mathrm{du} \\ $$$$ \\ $$ Answered by john santu last updated on 23/Mar/20 $$−\frac{\mathrm{1}}{\mathrm{2}}\int\:\frac{\mathrm{2}}{\mathrm{3}\left(\mathrm{2}{u}+\mathrm{1}\right)}{du}+\int\:\frac{\mathrm{5}}{\mathrm{3}\left({u}−\mathrm{1}\right)}{du} \\…
Question Number 85590 by sahnaz last updated on 23/Mar/20 $$\int\frac{\mathrm{1}+\mathrm{4u}}{−\mathrm{4u}^{\mathrm{2}} +\mathrm{2u}+\mathrm{2}}\mathrm{du} \\ $$$$ \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 85589 by sahnaz last updated on 23/Mar/20 $$\int\frac{\mathrm{1}+\mathrm{4u}}{−\mathrm{4u}^{\mathrm{2}} +\mathrm{2u}+\mathrm{2}}\mathrm{du} \\ $$$$ \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 151117 by mnjuly1970 last updated on 18/Aug/21 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 85568 by jagoll last updated on 23/Mar/20 $$\int\underset{\mathrm{0}} {\overset{\mathrm{2}\pi} {\:}}\:\frac{\mathrm{dx}}{\:\sqrt{\mathrm{2}}−\mathrm{cos}\:\mathrm{x}} \\ $$ Commented by jagoll last updated on 23/Mar/20 $$\mathrm{I}\:=\:\int\underset{\mathrm{0}} {\overset{\mathrm{2}\pi} {\:}}\:\frac{\mathrm{dx}}{\:\sqrt{\mathrm{2}}−\mathrm{cos}\:\mathrm{x}} \\…
Question Number 85551 by jagoll last updated on 22/Mar/20 $$\int\:\frac{\mathrm{dx}}{\mathrm{x}^{\mathrm{2}} \left(\mathrm{x}^{\mathrm{4}} +\mathrm{1}\right)^{\frac{\mathrm{3}}{\mathrm{4}}} } \\ $$ Commented by jagoll last updated on 23/Mar/20 Terms of Service…
Question Number 151069 by qaz last updated on 18/Aug/21 $$\mathrm{Calculate}\:\:::\:\:\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{sin}\:\left(\mathrm{1145141919810893x}\right)}{\mathrm{x}\left(\mathrm{cosh}\:\mathrm{x}+\mathrm{cos}\:\mathrm{x}\right)}\mathrm{dx}=\frac{\pi}{\mathrm{4}} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com