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Question Number 107107 by mohammad17 last updated on 08/Aug/20 Commented by Dwaipayan Shikari last updated on 08/Aug/20 $$\int_{\mathrm{0}} ^{\infty} \sqrt{\mathrm{y}}\:\:\mathrm{e}^{−\mathrm{t}^{\mathrm{2}} } \:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{y}^{\mathrm{3}} =\mathrm{t}^{\mathrm{2}\:\:} \:,\mathrm{y}^{\frac{\mathrm{3}}{\mathrm{2}}} =\mathrm{t}\:\:\:\:\:\frac{\mathrm{3}}{\mathrm{2}}\sqrt{\mathrm{y}}\:\:=\frac{\mathrm{dt}}{\mathrm{dy}}…
Question Number 172636 by Mikenice last updated on 29/Jun/22 Answered by floor(10²Eta[1]) last updated on 29/Jun/22 $$\mathrm{I}=\int\frac{\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{1}}−\sqrt{\mathrm{x}^{\mathrm{2}} −\mathrm{1}}}{\:\sqrt{\mathrm{x}^{\mathrm{2}} −\mathrm{1}}\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{1}}}\mathrm{dx}=\int\frac{\mathrm{dx}}{\:\sqrt{\mathrm{x}^{\mathrm{2}} −\mathrm{1}}}−\int\frac{\mathrm{dx}}{\:\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{1}}}=\mathrm{I}_{\mathrm{1}} −\mathrm{I}_{\mathrm{2}} \\…
Question Number 41561 by Tawa1 last updated on 09/Aug/18 $$\int\:\frac{\mathrm{dx}}{\mathrm{3sin}\left(\mathrm{x}\right)\:+\:\mathrm{4cos}\left(\mathrm{x}\right)} \\ $$ Commented by math khazana by abdo last updated on 09/Aug/18 $${changement}\:{tan}\left(\frac{{x}}{\mathrm{2}}\right)={t}\:{give} \\ $$$${I}\:\:=\:\int\:\:\:\:\:\:\frac{\mathrm{1}}{\mathrm{3}\frac{\mathrm{2}{t}}{\mathrm{1}+{t}^{\mathrm{2}}…
Question Number 172635 by Mikenice last updated on 29/Jun/22 Answered by MJS_new last updated on 29/Jun/22 $$\int\frac{\mathrm{4}^{{x}} }{\mathrm{4}^{{x}} +\mathrm{1}}{dx}= \\ $$$$\:\:\:\:\:\left[{t}=\mathrm{4}^{{x}} +\mathrm{1}\:\rightarrow\:{dx}=\frac{{dt}}{\mathrm{4}^{{x}} \:\mathrm{ln}\:\mathrm{4}}\right] \\ $$$$=\frac{\mathrm{1}}{\mathrm{ln}\:\mathrm{4}}\int\frac{{dt}}{{t}}=\frac{\mathrm{ln}\:{t}}{\mathrm{ln}\:\mathrm{4}}=…
Question Number 41555 by Raj Singh last updated on 09/Aug/18 Answered by alex041103 last updated on 09/Aug/18 $$\mathrm{1}.\:\int\frac{{ln}\:{x}}{\left(\mathrm{1}+{ln}\:{x}\right)^{\mathrm{2}} }\:{dx} \\ $$$${let}\:{x}={e}^{{u}} \:{dx}={e}^{{u}} {du}. \\ $$$$\int\frac{\mathrm{1}+{u}−\mathrm{1}}{\left(\mathrm{1}+{u}\right)^{\mathrm{2}}…
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