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Category: Integration

Question-152533

Question Number 152533 by mnjuly1970 last updated on 29/Aug/21 Answered by Kamel last updated on 29/Aug/21 $${I}\overset{{t}=\sqrt{{x}}} {=}\mathrm{2}\int_{\mathrm{0}} ^{+\infty} \frac{{Ln}\left(\mathrm{1}+{t}\right)}{{t}\left(\mathrm{1}+{t}\right)}{dt}=\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{Ln}\left(\mathrm{1}+{t}\right)}{{t}\left(\mathrm{1}+{t}\right)}{dt}+\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{Ln}\left(\mathrm{1}+{t}\right)−{Ln}\left({t}\right)}{\mathrm{1}+{t}}{dt} \

Question-86993

Question Number 86993 by Power last updated on 01/Apr/20 Commented by abdomathmax last updated on 01/Apr/20 I=15[10x]dxvhangement10x=tgive$${I}\:=\frac{\mathrm{1}}{\mathrm{10}}\:\int_{\mathrm{10}} ^{\mathrm{50}} \left[{t}\right]{dt}\:=\frac{\mathrm{1}}{\mathrm{10}}\sum_{{k}=\mathrm{10}} ^{\mathrm{49}} \:\int_{{k}}…

A-particle-is-projected-upwards-with-a-velocity-of-96ms-1-In-addition-to-being-subject-to-gravity-it-is-acted-on-by-a-retardation-of-16t-where-t-is-the-time-from-the-start-of-the-motion-

Question Number 152492 by nadovic last updated on 28/Aug/21 Aparticleisprojectedupwardswithavelocityof96ms1.Inadditiontobeingsubjecttogravity,itisactedonbyaretardationof16t,wheretisthetimefromthestartofthemotion.Whatisthegreatestheightattainedbytheparticle?