Question Number 226561 by Spillover last updated on 04/Dec/25 Answered by Spillover last updated on 04/Dec/25 Answered by Spillover last updated on 04/Dec/25 Answered by…
Question Number 226554 by Spillover last updated on 03/Dec/25 Answered by peace2 last updated on 03/Dec/25 $$=\frac{\mathrm{1}}{{n}}\int\frac{{nx}^{{n}−\mathrm{1}} {dx}}{{x}^{{n}} \left(\mathrm{1}+{x}^{{n}} \right)}=\frac{\mathrm{1}}{{n}}\int\frac{{dy}}{{y}\left(\mathrm{1}+{y}\right)}=\frac{\mathrm{1}}{{n}}\int\frac{\mathrm{1}}{{y}}−\frac{\mathrm{1}}{\mathrm{1}+{y}}{dy} \\ $$$$=\frac{\mathrm{1}}{{n}}{ln}\left(\frac{{y}}{\mathrm{1}+{y}}\right)+\mathrm{c};{y}={x}^{{n}} \\ $$ Answered…
Question Number 226292 by Spillover last updated on 25/Nov/25 Answered by Frix last updated on 25/Nov/25 $$=\int\frac{\mathrm{cos}\:{x}\:\mathrm{sin}\:{x}}{\mathrm{cos}\:{x}\:+\mathrm{sin}\:{x}}{dx}\:\overset{\left[{t}=\mathrm{tan}\:\frac{{x}}{\mathrm{2}}\right]} {=} \\ $$$$=\mathrm{4}\int\frac{{t}\left({t}^{\mathrm{2}} −\mathrm{1}\right)}{\left({t}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} \left({t}^{\mathrm{2}} −\mathrm{2}{t}−\mathrm{1}\right)}{dt}\:\overset{\left[\mathrm{decompose}\:\mathrm{etc}.\right]} {=}…
Question Number 226290 by Spillover last updated on 25/Nov/25 Answered by Frix last updated on 25/Nov/25 $$=\int\sqrt{\frac{\mathrm{cos}\:{x}\:\left(\mathrm{1}−\mathrm{cos}^{\mathrm{2}} \:{x}\right)}{\mathrm{1}−\mathrm{cos}^{\mathrm{3}} \:{x}}}\:{dx}= \\ $$$$=\int\mathrm{sin}\:{x}\sqrt{\frac{\mathrm{cos}\:{x}}{\mathrm{1}−\mathrm{cos}^{\mathrm{3}} \:{x}}}\:{dx}\:\overset{\left[{t}=\mathrm{cos}^{\frac{\mathrm{3}}{\mathrm{2}}} \:{x}\right]} {=} \\…
Question Number 226339 by Spillover last updated on 25/Nov/25 Answered by Ghisom_ last updated on 26/Nov/25 $$=\int\frac{{x}^{\mathrm{5}} }{{x}^{\mathrm{12}} −\mathrm{1}}{dx}= \\ $$$$\:\:\:\:\:\left[{t}={x}^{\mathrm{6}} \right] \\ $$$$=\frac{\mathrm{1}}{\mathrm{6}}\int\frac{{dt}}{{t}^{\mathrm{2}} −\mathrm{1}}=\frac{\mathrm{1}}{\mathrm{12}}\mathrm{ln}\:\frac{{t}−\mathrm{1}}{{t}+\mathrm{1}}\:=…
Question Number 226178 by Spillover last updated on 22/Nov/25 Answered by mr W last updated on 22/Nov/25 $$=\int\frac{\mathrm{1}+\frac{\mathrm{1}}{{x}}}{{x}+\frac{\mathrm{1}}{{x}}}{dx} \\ $$$$=\int\frac{{x}+\mathrm{1}}{{x}^{\mathrm{2}} +\mathrm{1}}{dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{d}\left({x}^{\mathrm{2}} +\mathrm{1}\right)}{{x}^{\mathrm{2}} +\mathrm{1}}+\int\frac{\mathrm{1}}{{x}^{\mathrm{2}}…
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Question Number 226147 by Spillover last updated on 20/Nov/25 Answered by Simurdiera last updated on 20/Nov/25 $$\mathrm{1} \\ $$ Terms of Service Privacy Policy Contact:…
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Question Number 225994 by Spillover last updated on 17/Nov/25 Answered by Simurdiera last updated on 18/Nov/25 $${Soluci}\acute {{o}n} \\ $$$$\int_{\mathrm{1}/\mathrm{2}} ^{\:\mathrm{1}} \frac{\sqrt{\mathrm{1}\:−\:{x}^{\mathrm{2}} }}{{x}\centerdot\mathrm{sin}\left(\mathrm{arccos}\left({x}\right)\right)\:+\:\mathrm{arcsin}\left({x}\right)}\centerdot\frac{\sqrt{\mathrm{1}\:−\:{x}^{\mathrm{2}} }}{\:\sqrt{\mathrm{1}\:−\:{x}^{\mathrm{2}} }}\:{dx}…