Question Number 11800 by tawa last updated on 01/Apr/17 $$\mathrm{Find}\:\mathrm{the}\:\mathrm{sum} \\ $$$$\underset{\mathrm{n}=\mathrm{1}} {\overset{\infty} {\sum}}\:\frac{\mathrm{1}}{\left(\mathrm{n}\:+\:\mathrm{1}\right)\sqrt{\mathrm{n}}\:\:+\:\:\mathrm{n}\sqrt{\mathrm{n}\:+\:\mathrm{1}}} \\ $$ Answered by FilupS last updated on 01/Apr/17 $$\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)\sqrt{{n}}+{n}\sqrt{{n}+\mathrm{1}}}=\frac{\mathrm{1}}{\:\sqrt{{n}}}−\frac{\mathrm{1}}{\:\sqrt{{n}+\mathrm{1}}} \\…
Question Number 77335 by aliesam last updated on 05/Jan/20 Commented by aliesam last updated on 05/Jan/20 Commented by msup trace by abdo last updated on…
Question Number 77323 by Boyka last updated on 05/Jan/20 Commented by Tony Lin last updated on 05/Jan/20 $$\int\frac{\mathrm{1}}{\:\sqrt{{cos}\left({x}−\frac{\pi}{\mathrm{2}}\right)}}{dx} \\ $$$$=\int\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}−\mathrm{2}{sin}^{\mathrm{2}} \left(\frac{\mathrm{2}{x}−\pi}{\mathrm{4}}\right)}}{dx} \\ $$$${let}\:{u}=\frac{\mathrm{2}{x}−\pi}{\mathrm{4}},\:\frac{{du}}{{dx}}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\mathrm{2}\int\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}−\mathrm{2}{sin}^{\mathrm{2}}…
Question Number 77290 by TawaTawa last updated on 05/Jan/20 $$\int\:\sqrt{\mathrm{x}^{\mathrm{3}} \:+\:\mathrm{x}^{\mathrm{4}} }\:\:\mathrm{dx} \\ $$ Commented by mathmax by abdo last updated on 05/Jan/20 $${let}\:{I}\:=\int\sqrt{{x}^{\mathrm{3}} \:+{x}^{\mathrm{4}}…
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Question Number 77269 by john santu last updated on 05/Jan/20 $${what}\:{is}\:\int\:\frac{\mathrm{1}}{\mathrm{tan}\:^{\mathrm{3}} \left({x}^{\mathrm{2}} −\mathrm{1}\right)}\:{dx}\:? \\ $$ Commented by MJS last updated on 05/Jan/20 $$\mathrm{I}\:\mathrm{don}'\mathrm{t}\:\mathrm{think}\:\mathrm{this}\:\mathrm{can}\:\mathrm{be}\:\mathrm{solved}\:\mathrm{at}\:\mathrm{all} \\ $$…
Question Number 142791 by mnjuly1970 last updated on 05/Jun/21 $$\:\:\:{Evaluate}::\:… \\ $$$$\:\:\:\:\:\:\:\Omega\::=\int_{\mathrm{0}} ^{\:\mathrm{1}} \:\frac{{li}_{\mathrm{2}} \left(\sqrt{{x}}\:\right)}{\mathrm{1}+\sqrt{{x}}}\:{dx}=?? \\ $$$$\:\:\:\:……….. \\ $$ Answered by Dwaipayan Shikari last updated…
Question Number 77242 by Tony Lin last updated on 04/Jan/20 $${prove}\:{that} \\ $$$$\:\int_{\mathrm{0}} ^{{a}} \sqrt{\mathrm{2}+\frac{{a}}{{x}}−\mathrm{2}\sqrt{\frac{{a}}{{x}}}\:}{dx}={a}\left[\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}{ln}\left(\sqrt{\mathrm{2}}+\mathrm{1}\right)+\mathrm{1}\right] \\ $$ Commented by mathmax by abdo last updated on…
Question Number 142763 by lapache last updated on 05/Jun/21 Commented by Ar Brandon last updated on 05/Jun/21 $$\mathcal{I}=\int\sqrt{\mathrm{tanx}}\mathrm{dx}\:,\:\mathrm{t}^{\mathrm{2}} =\mathrm{tanx}\:\Rightarrow\mathrm{2tdt}=\left(\mathrm{1}+\mathrm{t}^{\mathrm{4}} \right)\mathrm{dx} \\ $$$$\:\:\:=\int\frac{\mathrm{2t}^{\mathrm{2}} \mathrm{dt}}{\mathrm{1}+\mathrm{t}^{\mathrm{4}} }=\int\frac{\left(\mathrm{t}^{\mathrm{2}} +\mathrm{1}\right)+\left(\mathrm{t}^{\mathrm{2}}…
Question Number 77221 by john santu last updated on 04/Jan/20 $$\int\:\frac{\sqrt{\mathrm{2}+{x}}\:{dx}}{\:\sqrt{{x}^{\mathrm{5}} }}\:=\:? \\ $$ Answered by jagoll last updated on 04/Jan/20 $$\int\:\frac{\sqrt{\mathrm{2}+\mathrm{x}}}{\mathrm{x}^{\mathrm{2}} \sqrt{\mathrm{x}}}\mathrm{dx}\:=\:\int\:\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }\sqrt{\frac{\mathrm{2}}{\mathrm{x}}+\mathrm{1}}\:\mathrm{dx} \\…