Question Number 137474 by mnjuly1970 last updated on 03/Apr/21 $$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:…..{nice}\:\:……..\:\:{calculus}….. \\ $$$$\:\:\:\:\:\:\:\boldsymbol{\phi}=\int_{\mathrm{0}} ^{\:\mathrm{1}} {ln}\left(\sqrt[{\mathrm{3}}]{{ln}\left(\sqrt{\mathrm{1}−{x}}\right)}\:\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:{Im}\left(\boldsymbol{\phi}\right)=??? \\ $$ Answered by mindispower last updated on 03/Apr/21…
Question Number 137461 by mathlove last updated on 03/Apr/21 Answered by Ñï= last updated on 03/Apr/21 $$\int\frac{\mathrm{1}+{x}^{\mathrm{2}} }{\mathrm{2}{x}+{x}^{\mathrm{3}} }{dx}=\frac{\mathrm{1}}{\mathrm{3}}\int\frac{\mathrm{2}+\mathrm{3}{x}^{\mathrm{2}} +\mathrm{1}}{\mathrm{2}{x}+{x}^{\mathrm{3}} }{dx}=\frac{\mathrm{1}}{\mathrm{3}}{ln}\mid\mathrm{2}{x}+{x}^{\mathrm{3}} \mid+\frac{\mathrm{1}}{\mathrm{3}}\int\frac{{dx}}{{x}\left(\mathrm{2}+{x}^{\mathrm{2}} \right)} \\ $$$$=\frac{\mathrm{1}}{\mathrm{3}}{ln}\mid\mathrm{2}{x}+{x}^{\mathrm{3}}…
Question Number 137466 by Lordose last updated on 03/Apr/21 $$\Omega\:=\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{\mathrm{xtan}^{−\mathrm{1}} \left(\mathrm{x}\right)\mathrm{log}\left(\mathrm{x}\right)}{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }\mathrm{dx} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 6390 by FilupSmith last updated on 25/Jun/16 $$\mathrm{Solve}: \\ $$$${I}\left({n}\right)=\int_{\mathrm{0}} ^{\:{n}} \left(−\mathrm{1}\right)^{\lfloor{x}\rfloor} {x}^{\mathrm{2}} {dx} \\ $$ Commented by prakash jain last updated on…
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Question Number 137448 by liberty last updated on 03/Apr/21 $$\int_{\mathrm{0}} ^{\:\mathrm{3}/\mathrm{4}} \frac{{dx}}{\left({x}+\mathrm{1}\right)\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}}\:? \\ $$ Answered by MJS_new last updated on 03/Apr/21 $$\int\frac{{dx}}{\left({x}+\mathrm{1}\right)\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}}= \\…
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Question Number 137439 by mnjuly1970 last updated on 02/Apr/21 $$\:\:\:\:\:\:\:\:\:\:\:\:……{nice}\:\:{calculus}….. \\ $$$$\:\:\:\:{prove}\:{that}:: \\ $$$$\:\:\:\:\:\:\boldsymbol{\chi}=\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{{ln}\left({x}+\sqrt{\mathrm{1}−{x}^{\mathrm{2}} \:}\:\right)}{{x}}{dx}=\frac{\pi^{\mathrm{2}} }{\mathrm{16}}\:…. \\ $$ Answered by mindispower last updated…
Question Number 137420 by mnjuly1970 last updated on 02/Apr/21 $$\:\:\:\:\:\:\:\:\:\:\:\:\:……{mathematical}\:…\:…\:…\:{analysis}\left({II}\right)….. \\ $$$$\:\:\:\:\:\:\:{prove}\:\:{that}\::: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\Omega=\int_{\:\mathbb{R}} \left(\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−{x}^{\mathrm{2}} \right)^{{n}} }{\left({n}!\right)^{\mathrm{2}} }\right){dx}=\mathrm{1} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:…………………….. \\ $$ Commented…
Question Number 137419 by mnjuly1970 last updated on 02/Apr/21 $$\:\:\:\:\:\:\:………{mathematical}\:\:\:\:….\:\:\:{analysis}…….. \\ $$$$\:\:\:\:\:\:\:{evaluate}…. \\ $$$$\:\:\:\:\:\:\:\boldsymbol{\phi}=\int_{\mathrm{0}} ^{\:\infty} \frac{{e}^{\mathrm{2}\pi{x}} −{e}^{\pi{x}} }{{x}\left(\mathrm{1}+{e}^{\mathrm{2}\pi{x}} \right)\left(\mathrm{1}+{e}^{\pi{x}} \right)}{dx}=\lambda\int_{\mathrm{0}} ^{\:\mathrm{1}} {ln}\left(\Gamma\left({x}\right){dx}\right. \\ $$$$\:\:\:\:\:\:\:\:\:\:\lambda\:=\:??? \\…