Question Number 85637 by jagoll last updated on 23/Mar/20 $$\int\:\frac{\mathrm{dx}}{\mathrm{x}+\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{1}}}\: \\ $$ Commented by abdomathmax last updated on 23/Mar/20 $${I}\:=\int\:\:\frac{{dx}}{{x}+\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}}\:\:{we}\:{do}\:{the}\:{changement}\:{x}={sh}\left({t}\right)\:\Rightarrow \\ $$$${I}\:=\int\:\:\frac{{cht}}{{sht}\:+{cht}}{dt}\:=\int\:\:\frac{{e}^{{t}} \:+{e}^{−{t}}…
Question Number 151144 by mnjuly1970 last updated on 18/Aug/21 Answered by qaz last updated on 18/Aug/21 $$\frac{\mathrm{1}}{\mathrm{n}}\int_{−\infty} ^{+\infty} \frac{\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{e}^{\mathrm{x}} \right)}{\mathrm{cosh}\:\left(\frac{\mathrm{x}}{\mathrm{n}}\right)}\mathrm{dx} \\ $$$$=\int_{−\infty} ^{+\infty} \frac{\mathrm{tan}^{−\mathrm{1}}…
Question Number 151145 by mnjuly1970 last updated on 18/Aug/21 Answered by qaz last updated on 18/Aug/21 $$\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{cosh}\:^{\mathrm{2}} \left(\mathrm{x}^{\mathrm{2}} \right)}\mathrm{dx} \\ $$$$=\mathrm{2}\int_{\mathrm{0}} ^{\infty}…
Question Number 85600 by M±th+et£s last updated on 23/Mar/20 $$\int\frac{{x}^{\mathrm{2}} }{\:\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}\:{dx} \\ $$ Commented by mathmax by abdo last updated on 23/Mar/20 $${I}\:=\int\:\frac{{x}^{\mathrm{2}} }{\:\sqrt{\mathrm{1}+{x}^{\mathrm{2}}…
Question Number 85603 by M±th+et£s last updated on 23/Mar/20 $${prove}\:{the}\:{relation} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{li}_{\mathrm{5}} \left(\sqrt[{\mathrm{5}}]{{x}}\right)}{\:\sqrt[{\mathrm{5}}]{{x}}}{dx}=\frac{\mathrm{5}}{\mathrm{4}}\left(\frac{\mathrm{25}}{\mathrm{3072}}−\frac{\zeta\left(\mathrm{2}\right)}{\mathrm{2}^{\mathrm{6}} }+\frac{\zeta\left(\mathrm{3}\right)}{\mathrm{2}^{\mathrm{4}} }−\frac{\zeta\left(\mathrm{4}\right)}{\mathrm{2}^{\mathrm{2}} }+\zeta\left(\mathrm{5}\right)\right) \\ $$ Terms of Service Privacy Policy…
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