Question Number 151182 by mnjuly1970 last updated on 18/Aug/21 $$ \\ $$$$\int_{\mathrm{0}} ^{\:\infty} \frac{{sin}\left(\mathrm{2}{x}\right){ln}\left({x}\right)}{{x}}\:{dx}=\:{m}.\int_{\mathrm{0}} ^{\:\infty} \frac{\:{ln}\left(\mathrm{1}+\mathrm{2}{x}+{x}^{\mathrm{2}} \right)}{{x}\left({ln}^{\mathrm{2}} \left({x}\right)+\:\pi^{\:\mathrm{2}} \right)}\:{dx} \\ $$$$\:\:\:\:\:\:\:\:\:\:{m}=?…. \\ $$ Terms of…
Question Number 85646 by M±th+et£s last updated on 23/Mar/20 $${show}\:{that} \\ $$$$\int\frac{\mathrm{1}}{\left[{x}\left({x}−\mathrm{1}\right)\left({x}−\mathrm{2}\right)\left({x}−\mathrm{3}\right)…\left({x}−{m}\right)\right]^{\mathrm{2}} }{dx}= \\ $$$$=\frac{\mathrm{1}}{\left({m}!\right)^{\mathrm{2}} }\underset{{n}=\mathrm{0}} {\overset{{m}} {\sum}}\frac{\begin{pmatrix}{{m}}\\{{n}}\end{pmatrix}^{\mathrm{2}} }{{n}−{x}}+\frac{\mathrm{2}}{\left({m}!\right)^{\mathrm{2}} }{ln}\mid\underset{{n}=\mathrm{0}} {\overset{{m}} {\prod}}\left({x}−{n}\right)^{\begin{pmatrix}{{m}}\\{{n}}\end{pmatrix}^{\mathrm{2}} \left({H}_{{m}−{n}} −{H}_{{n}} \right)}…
Question Number 85641 by abdomathmax last updated on 23/Mar/20 $${calculate}\:{A}_{\lambda} =\int_{\mathrm{3}} ^{\infty} \:\:\frac{{dx}}{\left({x}+\lambda\right)^{\mathrm{3}} \left({x}−\mathrm{2}\right)^{\mathrm{4}} }\:\:\:\left(\lambda>\mathrm{0}\right) \\ $$ Commented by mathmax by abdo last updated on…
Question Number 85637 by jagoll last updated on 23/Mar/20 $$\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} \\…