Question Number 55067 by gunawan last updated on 17/Feb/19 $$\mathrm{Known}\:\:{C}\:\mathrm{circle}\:\mathrm{centered}\:\mathrm{at}\:\mathrm{0}. \\ $$$$\mathrm{Find}\:\mathrm{value}\:\mathrm{than}\:\int_{{C}} \frac{{dz}}{\mathrm{1}−{z}} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 186132 by normans last updated on 01/Feb/23 Commented by MJS_new last updated on 01/Feb/23 $$\mathrm{I}\:\mathrm{will}\:\mathrm{solve}\:\mathrm{this}\:\mathrm{and}\:\mathrm{your}\:\mathrm{earlier}\:\mathrm{integral} \\ $$$$\mathrm{immediately}\:\mathrm{after}\:\mathrm{I}\:\mathrm{finished}\:\mathrm{my}\:\mathrm{proof}\:\mathrm{of} \\ $$$$\mathrm{Riemann}'\mathrm{s}\:\mathrm{Hypothesis}. \\ $$$$\mathrm{see}\:\mathrm{you}\:\mathrm{later}\:\mathrm{alligator}! \\ $$…
Question Number 120599 by MagdiRagheb last updated on 01/Nov/20 $${Let}\:{x}\:=\:{u}^{\mathrm{6}} \:\:\:\:\:\:\:\:\:{dx}\:=\:\mathrm{6}{u}^{\mathrm{5}} \:{du} \\ $$$${I}\:=\:\int\frac{{u}^{\mathrm{3}} }{\left(\mathrm{1}+{u}^{\mathrm{2}} \right)^{\mathrm{2}} }\:×\mathrm{6}{u}^{\mathrm{5}} \:{du} \\ $$$$\:\:\:=\mathrm{6}\:\int\frac{{u}^{\mathrm{8}} }{\mathrm{1}+\mathrm{2}{u}^{\mathrm{2}} +{u}^{\mathrm{4}} }\:{du} \\ $$$$\:\:\:=\mathrm{6}\:\int\left[\frac{−\mathrm{4}}{{u}^{\mathrm{2}}…
Question Number 55051 by maxmathsup by imad last updated on 16/Feb/19 $${calculate}\:{lim}_{{n}\rightarrow+\infty} \:\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{dx}}{\mathrm{1}+{x}+{x}^{\mathrm{2}} \:+…+{x}^{{n}} } \\ $$ Commented by maxmathsup by imad last…
Question Number 120582 by MagdiRagheb last updated on 01/Nov/20 $${Let}\:{u}={x}^{\frac{\mathrm{3}}{\mathrm{5}}} \:\:\:\:\:{du}\:=\:\frac{\mathrm{3}}{\mathrm{5}{x}^{\frac{\mathrm{2}}{\mathrm{5}}} } \\ $$$${I}\:=\:\frac{\mathrm{5}}{\mathrm{3}}\int\frac{{u}}{\:\sqrt{\mathrm{3}−\mathrm{2}{u}}}\:{du}\:=\:−\frac{\mathrm{5}}{\mathrm{6}}\:\int\frac{\mathrm{3}−\mathrm{2}{u}−\mathrm{3}}{\:\sqrt{\mathrm{3}−\mathrm{2}{u}}}\:{du} \\ $$$$=\:−\frac{\mathrm{5}}{\mathrm{3}}\:\int\left[\sqrt{\mathrm{3}−\mathrm{2}{u}}\:−\mathrm{3}\left(\mathrm{3}−\mathrm{2}{u}\right)^{−\frac{\mathrm{1}}{\mathrm{2}}} \right]\:{du} \\ $$$$=\:−\frac{\mathrm{5}}{\mathrm{3}}\left[−\frac{\mathrm{1}}{\mathrm{3}}\left(\mathrm{3}−\mathrm{2}{u}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} +\mathrm{3}\left(\mathrm{3}−\mathrm{2}{u}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} \right]+{c} \\ $$$$=\:−\frac{\mathrm{5}}{\mathrm{18}}\left(\mathrm{3}−\mathrm{2}{u}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} \left[−\mathrm{3}+\mathrm{2}{u}\:+\:\mathrm{9}\right]+{c} \\…
Question Number 55030 by Joel578 last updated on 16/Feb/19 $$\mathrm{Find} \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\left(\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\:\left(\mathrm{ln}\:{x}\right)^{{n}} \:{dx}\right) \\ $$ Commented by tm888 last updated on 16/Feb/19…
Question Number 120562 by peter frank last updated on 01/Nov/20 Answered by Dwaipayan Shikari last updated on 01/Nov/20 $$\int\mathrm{1}−\frac{{bx}+\mathrm{12}}{{x}^{\mathrm{2}} +{bx}+\mathrm{12}}{dx} \\ $$$$={x}−{b}\int\frac{{x}+\frac{\mathrm{12}}{{b}}}{{x}^{\mathrm{2}} +{bx}+\mathrm{12}}{dx} \\ $$$$={x}−\frac{{b}}{\mathrm{2}}\int\frac{\mathrm{2}{x}+{b}}{{x}^{\mathrm{2}}…
Question Number 120554 by snipers237 last updated on 01/Nov/20 $${Prove}\:{that}\:{for}\:{all}\:\:{a}>\mathrm{0} \\ $$$$\int_{\left[−{a};{a}\right]} {arg}\left(\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}\:−{ix}\right)\right){dx}\:=\mathrm{0} \\ $$$${Deduce}\:{that}\: \\ $$$${f}:\:{x}\rightarrow{arg}\left(\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}\:−{ix}\right)\right)\:\:{is}\:{an}\:{old}\:{function}\:{on}\:\mathbb{R} \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 120552 by talminator2856791 last updated on 01/Nov/20 $$\: \\ $$$$\: \\ $$$$\:\:\:\:\:\:\:\mathrm{evaluate}:\:\int_{\mathrm{0}} ^{\:\infty} \left(\frac{{x}}{{x}+\mathrm{1}}\right)^{{x}!} {dx} \\ $$$$ \\ $$$$ \\ $$ Terms of…
Question Number 186086 by Rupesh123 last updated on 31/Jan/23 Terms of Service Privacy Policy Contact: info@tinkutara.com