Question Number 227271 by Spillover last updated on 11/Jan/26 Answered by Kassista last updated on 11/Jan/26 $$ \\ $$$$\int\:\frac{{x}^{\mathrm{2}} }{{x}^{\mathrm{2}} +\mathrm{5}}\:{dx}\:=\:\int\:\frac{{x}^{\mathrm{2}} +\mathrm{5}−\mathrm{5}}{{x}^{\mathrm{2}} +\mathrm{5}}\:{dx}\:=\:\int\:\frac{{x}^{\mathrm{2}} +\mathrm{5}}{{x}^{\mathrm{2}} +\mathrm{5}}\:{dx}\:−\int\frac{\mathrm{5}}{{x}^{\mathrm{2}}…
Question Number 227255 by Spillover last updated on 11/Jan/26 Answered by Spillover last updated on 11/Jan/26 Answered by Spillover last updated on 11/Jan/26 Answered by…
Question Number 227184 by Spillover last updated on 04/Jan/26 Answered by Spillover last updated on 04/Jan/26 Answered by Spillover last updated on 04/Jan/26 Answered by…
Question Number 227149 by Spillover last updated on 03/Jan/26 Answered by Kassista last updated on 03/Jan/26 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 227146 by Spillover last updated on 03/Jan/26 Answered by Kassista last updated on 03/Jan/26 $$ \\ $$$${I}\:=\:\int_{\:−\mathrm{2}} ^{\:\mathrm{2}\:} \frac{{x}^{\mathrm{2}} }{\mathrm{1}+\mathrm{5}^{{x}} }\:{dx}\:\overset{{x}=−{u}} {\Rightarrow}\int_{\:\mathrm{2}} ^{\:−\mathrm{2}}…
Question Number 227128 by Spillover last updated on 01/Jan/26 Answered by som(math1967) last updated on 01/Jan/26 $$\int_{\mathrm{0}\:\:} ^{\frac{\pi}{\mathrm{2}}} \frac{\sqrt{{tanx}}}{\mathrm{1}+\sqrt{{tanx}}}{dx} \\ $$$${I}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\sqrt{{tan}\left(\frac{\pi}{\mathrm{2}}−{x}\right)}{dx}}{\mathrm{1}+\sqrt{{tan}\left(\frac{\pi}{\mathrm{2}}−{x}\right)}} \\ $$$${I}=\int_{\mathrm{0}}…
Question Number 227113 by MrAjder last updated on 01/Jan/26 $$\int_{\mathrm{0}} ^{\mathrm{1}} \lfloor\mathrm{log}_{\mathrm{2}} \left({x}−\mathrm{2}^{\lfloor\mathrm{log}_{\mathrm{2}} {x}\rfloor} \right)\rfloor{dx} \\ $$ Answered by MrAjder last updated on 01/Jan/26 Terms…
Question Number 227131 by Devil001 last updated on 01/Jan/26 $$\int \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 227127 by MrAjder last updated on 01/Jan/26 $$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{prove}: \\ $$$$\:\:\:\:\:\:\:\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{ln}{x}}{\mathrm{1}+{x}^{\mathrm{6}} }{dx}=\frac{\pi^{\mathrm{2}} }{\mathrm{12}\sqrt{\mathrm{3}}}−\frac{\mathrm{5}}{\mathrm{9}}\boldsymbol{\mathrm{G}} \\ $$$$ \\ $$ Commented by Tawa11 last updated…
Question Number 227108 by gregori last updated on 31/Dec/25 $$\:\underset{\mathrm{0}} {\overset{\pi} {\int}}\:\frac{\mathrm{sin}\:\mathrm{2}{x}}{\mathrm{3}}\sqrt{\mathrm{cos}\:{x}+\mathrm{1}}\:{dx}\:=? \\ $$ Answered by Frix last updated on 31/Dec/25 $$=\frac{\mathrm{2}}{\mathrm{3}}\underset{\mathrm{0}} {\overset{\pi} {\int}}\mathrm{cos}\:{x}\:\mathrm{sin}\:{x}\:\sqrt{\mathrm{1}+\mathrm{cos}\:{x}}\:{dx}\:\overset{\left[{t}=\sqrt{\mathrm{1}+\mathrm{cos}\:{x}}\right]} {=}…