Question Number 147948 by tabata last updated on 24/Jul/21 Commented by tabata last updated on 24/Jul/21 $${Msr}\:{olaf}\:{help}\:{me}\:{please}\:? \\ $$ Answered by Olaf_Thorendsen last updated on…
Question Number 82410 by naka3546 last updated on 21/Feb/20 Commented by naka3546 last updated on 21/Feb/20 $${find}\:\:{real}\:\:{solution}\:\:{of}\:\:{the}\:\:{equation}\:\:{above} \\ $$ Answered by MJS last updated on…
Question Number 82392 by naka3546 last updated on 21/Feb/20 Commented by naka3546 last updated on 21/Feb/20 $${Area}\:\:{of}\:\:{shaded}\:\:{region}\:\:{is}\:\:… \\ $$ Commented by mr W last updated…
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Question Number 147918 by abdurehime last updated on 24/Jul/21 Commented by abdurehime last updated on 24/Jul/21 Commented by abdurehime last updated on 24/Jul/21 $$\mathrm{please}\:\mathrm{help}\:\mathrm{me} \\…
Question Number 147915 by tabata last updated on 24/Jul/21 Commented by tabata last updated on 24/Jul/21 $${how}\:{can}\:{solve}\:{this}\:{please}\:? \\ $$ Commented by tabata last updated on…
Question Number 82365 by M±th+et£s last updated on 20/Feb/20 Commented by mr W last updated on 21/Feb/20 $${a}_{{n}} =\mathrm{1}+\frac{\mathrm{3}\left({n}−\mathrm{1}\right){n}}{\mathrm{2}} \\ $$ Commented by mr W…
Question Number 147873 by 0731619 last updated on 24/Jul/21 Answered by Olaf_Thorendsen last updated on 24/Jul/21 $$\mathrm{S}\:=\:\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(\frac{\mathrm{1}}{\mathrm{3}{n}+\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{3}{n}+\mathrm{4}}\right) \\ $$$$\mathrm{S}\:=\:\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{2}}{\left(\mathrm{3}{n}+\mathrm{2}\right)\left(\mathrm{3}{n}+\mathrm{4}\right)} \\ $$$$\mathrm{S}\:=\:\frac{\mathrm{2}}{\mathrm{9}}\underset{{n}=\mathrm{0}}…
Question Number 147879 by 0731619 last updated on 24/Jul/21 Answered by puissant last updated on 24/Jul/21 $$\mathrm{I}=\int\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{5}} +\mathrm{1}}\mathrm{dx} \\ $$$$=\int\frac{\mathrm{dx}}{\left(\mathrm{x}+\mathrm{1}\right)\left(\mathrm{x}^{\mathrm{4}} −\mathrm{x}^{\mathrm{3}} +\mathrm{x}^{\mathrm{2}} −\mathrm{x}+\mathrm{1}\right)} \\ $$$$=\int\frac{\frac{\mathrm{1}}{\mathrm{5}}}{\mathrm{x}+\mathrm{1}}\mathrm{dx}+\int\frac{−\frac{\mathrm{1}}{\mathrm{5}}\mathrm{x}^{\mathrm{3}}…