Question Number 124261 by Ar Brandon last updated on 02/Dec/20 $$\int\frac{\mathrm{2x}^{\mathrm{2}} +\mathrm{5x}+\mathrm{9}}{\left(\mathrm{x}+\mathrm{1}\right)\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{x}+\mathrm{1}}}\mathrm{dx} \\ $$ Commented by MJS_new last updated on 02/Dec/20 $$\mathrm{I}\:\mathrm{think}\:\mathrm{something}\:\mathrm{went}\:\mathrm{wrong}… \\ $$…
Question Number 124252 by bramlexs22 last updated on 02/Dec/20 $$\:\:{o}\left({x}\right)=\int\:\frac{{dx}}{\mathrm{sec}\:^{\mathrm{3}} {x}\:\mathrm{sin}\:^{\mathrm{4}} {x}}\: \\ $$ Answered by Dwaipayan Shikari last updated on 02/Dec/20 $$\int\frac{{cos}^{\mathrm{3}} {x}}{{sin}^{\mathrm{4}} {x}}{dx}=\int\frac{{cosx}\left(\mathrm{1}−{sin}^{\mathrm{2}}…
Question Number 58717 by rahul 19 last updated on 28/Apr/19 Commented by rahul 19 last updated on 28/Apr/19 $${Curve}\:{is}\:{y}={e}^{−{x}^{\mathrm{2}} } . \\ $$$${Ans}:{a},{b},{d}. \\ $$…
Question Number 124251 by bramlexs22 last updated on 02/Dec/20 $$\:\underset{\mathrm{0}} {\overset{\infty} {\int}}\:\frac{{x}^{\mathrm{2}} }{\mathrm{cosh}\:{x}}\:{dx}\:? \\ $$ Commented by Dwaipayan Shikari last updated on 02/Dec/20 $$\mathrm{2}\underset{{n}=\mathrm{1}} {\overset{\infty}…
Question Number 124228 by mnjuly1970 last updated on 01/Dec/20 $$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:…:::\:\:{nice}\:\:{calculus}:::… \\ $$$$\:\:\:\:\:\:{evaluate} \\ $$$$\:\:\:\:\:\:\:\:\:\mathrm{I}=\int_{\mathrm{0}} ^{\:\infty} \frac{{cos}\left({ln}\left({x}\right)\right)}{\mathrm{1}+{x}^{\mathrm{3}} }{dx}=….??? \\ $$ Answered by Lordose last updated on…
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Question Number 189752 by TUN last updated on 21/Mar/23 $$\underset{\mathrm{0}} {\int}^{\infty} {x}^{\mathrm{2}} .{e}^{−{x}^{\mathrm{2}} } {dx}=¿ \\ $$ Answered by MJS_new last updated on 21/Mar/23 $$\int{x}^{\mathrm{2}}…
Question Number 124217 by bramlexs22 last updated on 01/Dec/20 $$\underset{\mathrm{0}} {\overset{\infty} {\int}}\:\frac{{dx}}{{x}\:\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}\:? \\ $$ Answered by Dwaipayan Shikari last updated on 01/Dec/20 $$\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\:={t}\Rightarrow\frac{{x}}{\:\sqrt{\mathrm{1}+{x}^{\mathrm{2}}…
Question Number 124202 by mnjuly1970 last updated on 01/Dec/20 $$\:\:\:\:\:\:\:\:\:\::::\:\:{nice}\:\:{calculus}\:::: \\ $$$$\:\:\:\:\:{prove}\:\:{that}\:::: \\ $$$$ \\ $$$$\underset{{m},{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left\{\frac{\left(−\mathrm{1}\right)^{{n}+{m}} }{{n}^{\mathrm{2}} +{m}^{\mathrm{2}} }\right\}\:\overset{???} {=}\frac{\pi^{\mathrm{2}} }{\mathrm{12}}\:−\frac{\pi}{\mathrm{4}}{ln}\left(\mathrm{2}\right) \\ $$…
Question Number 124201 by mnjuly1970 last updated on 01/Dec/20 $$\:\:\:\:\:\:\:\:\:\:\:\:\:\::::\:\:{nice}\:\:{calculus}\:::: \\ $$$$\:\:\:\:\:{please}\:\:{prove}\:::: \\ $$$$\:\:\:\:\Omega\:=\:\int_{\mathrm{0}} ^{\:\infty} \frac{{x}^{\frac{\mathrm{1}}{\mathrm{2}}} }{{x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{5}}{dx}=\frac{\pi}{\:\sqrt{\varphi}} \\ $$$$\:\:\:\:\:{where}\:\:\varphi\:\:{is}\:{Golden}\:{ratio}… \\ $$ Answered by ajfour…