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Category: Integration

advanced-calculus-lim-n-1-n-x-x-2-dx-n-1-solution-n-1-n-x-x-2-dx-k-1-n-1-k-k-1-x

Question Number 139556 by mnjuly1970 last updated on 28/Apr/21 $$\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:……..\:{advanced}\:…\:…\:…\:{calculus}…….. \\ $$$$\:\:\:\Phi=\:{lim}_{{n}\rightarrow\infty} \left\{\int_{\mathrm{1}} ^{\:{n}} \frac{{x}}{\left[{x}\right]^{\mathrm{2}} }\:{dx}\:−\psi\left({n}+\mathrm{1}\right)\right\}=? \\ $$$$\:\:\:\:{solution}: \\ $$$$\:\:\:\:\:\Phi_{{n}} =\int_{\mathrm{1}} ^{\:{n}} \frac{{x}}{\left[{x}\right]^{\mathrm{2}}…

advanced-calculus-prove-that-i-0-1-2-e-2x-1-1-e-x-1-x-dx-log-1-pi-ii-0-1-2-1-1-e-x-e-2x-x-dx-log-pi-2

Question Number 139524 by mnjuly1970 last updated on 28/Apr/21 $$\:\:\:\:\:\:\:\:\:\:\:\:……{advanced}\:\:{calculus}….. \\ $$$$\:\:\:\:\:\:\:\:\:{prove}\:\:{that}: \\ $$$$\:{i}::\boldsymbol{\phi}=\int_{\mathrm{0}} ^{\:\infty} \left(\frac{\mathrm{1}}{\mathrm{2}}{e}^{−\mathrm{2}{x}} −\frac{\mathrm{1}}{\mathrm{1}+{e}^{{x}} }\right)\frac{\mathrm{1}}{{x}}\:{dx}={log}\left(\frac{\mathrm{1}}{\:\sqrt{\pi}}\:\right) \\ $$$$\:{ii}::\int_{\mathrm{0}} ^{\:\infty} \left(\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{1}+{e}^{−{x}} }\right)\frac{{e}^{−\mathrm{2}{x}} }{{x}}{dx}={log}\left(\frac{\sqrt{\pi}}{\mathrm{2}}\right) \\…

Question-139476

Question Number 139476 by aliibrahim1 last updated on 27/Apr/21 Answered by qaz last updated on 27/Apr/21 $${I}=\int\frac{{dx}}{{x}\left({x}+\mathrm{1}\right)\left({x}+\mathrm{2}\right)…\left({x}+{n}\right)} \\ $$$${f}\left({x}\right)=\frac{\mathrm{1}}{{x}\left({x}+\mathrm{1}\right)\left({x}+\mathrm{2}\right)…\left({x}+{n}\right)}=\frac{{A}_{\mathrm{0}} }{{x}}+\frac{{A}_{\mathrm{1}} }{{x}+\mathrm{1}}+\frac{{A}_{\mathrm{2}} }{{x}+\mathrm{2}}+…+\frac{{A}_{{n}} }{{x}+{n}} \\ $$$${A}_{\mathrm{0}}…

nice-calculus-0-1-ln-1-x-2-x-2-1-dx-pi-2-24-NOTE-li-2-z-li-2-1-z-pi-2-6-ln-z-ln-1-z-Hence-

Question Number 139478 by mnjuly1970 last updated on 27/Apr/21 $$\:\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:………\:{nice}\:…\:…\:…\:{calculus}…….. \\ $$$$\:\:\:\:\:\:\Phi:=\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{{ln}\left(\frac{\mathrm{1}+{x}}{\mathrm{2}}\right)}{{x}^{\mathrm{2}} −\mathrm{1}}{dx}=\frac{\pi^{\mathrm{2}} }{\mathrm{24}} \\ $$$$\:\:\:\:\:\:\:{NOTE}\:::\:{li}_{\mathrm{2}} \left({z}\right)+{li}_{\mathrm{2}} \left(\mathrm{1}−{z}\right)=\frac{\pi^{\mathrm{2}} }{\mathrm{6}}−{ln}\left({z}\right){ln}\left(\mathrm{1}−{z}\right) \\ $$$$\:\:\:\:\:\:\:\:\:{Hence}\:::\:\:{li}_{\mathrm{2}}…

Question-139454

Question Number 139454 by aliibrahim1 last updated on 27/Apr/21 Answered by Dwaipayan Shikari last updated on 27/Apr/21 $$\int{x}^{\mathrm{2}} \frac{{tan}^{−\mathrm{1}} {x}}{{x}^{\mathrm{2}} +\mathrm{1}}{dx} \\ $$$$=\int{tan}^{−\mathrm{1}} \left({x}\right)−\int\frac{{tan}^{−\mathrm{1}} \left({x}\right)}{{x}^{\mathrm{2}}…

Evaluate-0-1-2-1-x-2-dx-By-direct-integration-and-by-expanding-as-a-power-series-

Question Number 8375 by tawakalitu last updated on 09/Oct/16 $$\mathrm{Evaluate}\::\:\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \sqrt{\mathrm{1}\:−\:\mathrm{x}^{\mathrm{2}} }\:\mathrm{dx} \\ $$$$\mathrm{By}\:\mathrm{direct}\:\mathrm{integration}\:\mathrm{and}\:\mathrm{by}\:\mathrm{expanding} \\ $$$$\mathrm{as}\:\mathrm{a}\:\mathrm{power}\:\mathrm{series}. \\ $$ Terms of Service Privacy Policy Contact:…