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Category: Integration

Question-74720

Question Number 74720 by chess1 last updated on 29/Nov/19 Commented by mathmax by abdo last updated on 29/Nov/19 $${changement}\:\sqrt{\frac{{x}−\mathrm{1}}{{x}+\mathrm{1}}}\:={t}\:{give}\:{x}−\mathrm{1}\:={t}^{\mathrm{2}} \left({x}+\mathrm{1}\right)\:\Rightarrow\left(\mathrm{1}−{t}^{\mathrm{2}} \right){x}=\mathrm{1}+{t}^{\mathrm{2}} \:\Rightarrow \\ $$$${x}=\frac{\mathrm{1}+{t}^{\mathrm{2}} }{\mathrm{1}−{t}^{\mathrm{2}}…

x-x-dx-

Question Number 9158 by geovane10math last updated on 21/Nov/16 $$\int{x}^{{x}} \:{dx}\:=\:? \\ $$ Answered by FilupSmith last updated on 22/Nov/16 $${x}^{{x}} ={e}^{{x}\mathrm{ln}\left({x}\right)} \\ $$$${e}^{{x}\mathrm{ln}\left({x}\right)} =\underset{{n}=\mathrm{0}}…

please-integrate-f-x-0-1-1-z-log-z-2-2zcos-x-1-z-1-2-dz-

Question Number 140194 by mnjuly1970 last updated on 05/May/21 $$\:\: \\ $$$$\:\:\:{please}\:\:{integrate}:: \\ $$$$\:\:\:\:\:{f}\left({x}\right)=\int_{\mathrm{0}} ^{\:\mathrm{1}} \left\{\frac{\mathrm{1}}{{z}}{log}\left(\frac{{z}^{\mathrm{2}} +\mathrm{2}{zcos}\left({x}\right)+\mathrm{1}}{\left({z}+\mathrm{1}\right)^{\mathrm{2}} }\right)\right\}{dz} \\ $$$$ \\ $$ Answered by Dwaipayan…

Question-9107

Question Number 9107 by jainamanj98@gmail.com last updated on 18/Nov/16 Answered by FilupSmith last updated on 19/Nov/16 $$\left(\mathrm{1}\right) \\ $$$$\int_{\mathrm{1}} ^{\mathrm{3}} \left({x}+\mathrm{3}\sqrt{{x}}\right){dx}=\int_{\mathrm{1}} ^{\mathrm{3}} {xdx}+\mathrm{3}\int_{\mathrm{1}} ^{\mathrm{3}} {x}^{\mathrm{1}/\mathrm{2}}…

1-calculate-f-x-0-1-t-2-x-2-t-2-dt-with-x-gt-0-2-calculste-g-x-0-1-t-2-x-2-t-2-dt-

Question Number 74637 by mathmax by abdo last updated on 28/Nov/19 $$\left.\mathrm{1}\right){calculate}\:{f}\left({x}\right)=\int_{\mathrm{0}} ^{\mathrm{1}} {t}^{\mathrm{2}} \sqrt{{x}^{\mathrm{2}} +{t}^{\mathrm{2}} }{dt}\:\:{with}\:{x}>\mathrm{0} \\ $$$$\left.\mathrm{2}\right)\:{calculste}\:{g}\left({x}\right)=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{t}^{\mathrm{2}} }{\:\sqrt{{x}^{\mathrm{2}} +{t}^{\mathrm{2}} }}{dt} \\…