Question Number 152533 by mnjuly1970 last updated on 29/Aug/21 Answered by Kamel last updated on 29/Aug/21 $${I}\overset{{t}=\sqrt{{x}}} {=}\mathrm{2}\int_{\mathrm{0}} ^{+\infty} \frac{{Ln}\left(\mathrm{1}+{t}\right)}{{t}\left(\mathrm{1}+{t}\right)}{dt}=\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{Ln}\left(\mathrm{1}+{t}\right)}{{t}\left(\mathrm{1}+{t}\right)}{dt}+\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{Ln}\left(\mathrm{1}+{t}\right)−{Ln}\left({t}\right)}{\mathrm{1}+{t}}{dt} \\…
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Question Number 86993 by Power last updated on 01/Apr/20 Commented by abdomathmax last updated on 01/Apr/20 $${I}\:=\int_{\mathrm{1}} ^{\mathrm{5}} \left[\mathrm{10}{x}\right]{dx}\:\:{vhangement}\:\mathrm{10}{x}\:={t}\:{give} \\ $$$${I}\:=\frac{\mathrm{1}}{\mathrm{10}}\:\int_{\mathrm{10}} ^{\mathrm{50}} \left[{t}\right]{dt}\:=\frac{\mathrm{1}}{\mathrm{10}}\sum_{{k}=\mathrm{10}} ^{\mathrm{49}} \:\int_{{k}}…
Question Number 86995 by M±th+et£s last updated on 01/Apr/20 $$\int_{\mathrm{0}} ^{\pi} \frac{{a}^{{n}} {sin}^{\mathrm{2}} \left({x}\right)+{b}^{{n}} {cos}^{\mathrm{2}} \left({x}\right)}{{a}^{\mathrm{2}{n}} {sin}^{\mathrm{2}} \left({x}\right)+{b}^{\mathrm{2}{n}} {cos}^{\mathrm{2}} \left({x}\right)}{dx}\:;\:{a}>{b} \\ $$ Answered by TANMAY…
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Question Number 152502 by Tawa11 last updated on 29/Aug/21 $$\int\:\mathrm{9x}^{\mathrm{2}} \left(\mathrm{4x}^{\mathrm{2}} \:\:+\:\:\mathrm{3}\right)^{\mathrm{10}} \:\mathrm{dx} \\ $$ Answered by Olaf_Thorendsen last updated on 30/Aug/21 $$\mathrm{F}\left({x}\right)\:=\:\int\mathrm{9}{x}^{\mathrm{2}} \left(\mathrm{4}{x}^{\mathrm{2}} +\mathrm{3}\right)^{\mathrm{10}}…
Question Number 86963 by abdomathmax last updated on 01/Apr/20 $${calculate}\:\:{I}\:=\int\:\:{cos}^{\mathrm{4}} {x}\:{sh}^{\mathrm{2}} {x}\:{dx} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 152492 by nadovic last updated on 28/Aug/21 $$\:\mathrm{A}\:\mathrm{particle}\:\mathrm{is}\:\mathrm{projected}\:\mathrm{upwards}\:\mathrm{with} \\ $$$$\:\mathrm{a}\:\mathrm{velocity}\:\mathrm{of}\:\:\mathrm{96}{ms}^{−\mathrm{1}} .\:\mathrm{In}\:\mathrm{addition}\:\mathrm{to} \\ $$$$\:\mathrm{being}\:\mathrm{subject}\:\mathrm{to}\:\mathrm{gravity},\:\mathrm{it}\:\mathrm{is}\:\mathrm{acted}\:\mathrm{on} \\ $$$$\:\mathrm{by}\:\mathrm{a}\:\mathrm{retardation}\:\mathrm{of}\:\mathrm{16}{t},\:\mathrm{where}\:{t}\:\mathrm{is}\:\mathrm{the} \\ $$$$\:\mathrm{time}\:\mathrm{from}\:\mathrm{the}\:\mathrm{start}\:\mathrm{of}\:\mathrm{the}\:\mathrm{motion}. \\ $$$$\:\mathrm{What}\:\mathrm{is}\:\mathrm{the}\:\mathrm{greatest}\:\mathrm{height}\:\mathrm{attained} \\ $$$$\:\mathrm{by}\:\mathrm{the}\:\mathrm{particle}? \\ $$…
Question Number 86957 by abdomathmax last updated on 01/Apr/20 $${find}\:\int\:\:{arctan}\left(\frac{\mathrm{1}−{u}}{\mathrm{1}+{u}}\right){du} \\ $$ Commented by Ar Brandon last updated on 01/Apr/20 $${I}=\int{arctan}\left(\frac{\mathrm{1}−{x}}{\mathrm{1}+{x}}\right){dx} \\ $$$${Let}\:{u}={arctan}\left(\frac{\mathrm{1}−{x}}{\mathrm{1}+{x}}\right)\:\Leftrightarrow\:{u}={arctan}\left({t}\right)\:\:{with}\:\:{t}=\frac{\mathrm{1}−{x}}{\mathrm{1}+{x}} \\ $$$$\Rightarrow\frac{{du}}{{dx}}=\frac{{du}}{{dt}}×\frac{{dt}}{{du}}…
Question Number 152494 by mnjuly1970 last updated on 28/Aug/21 $$ \\ $$$$\:\:\:\:{prove}\:{that}\::: \\ $$$$\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\Omega\::=\:\int_{\mathrm{0}\:} ^{\:\infty} \frac{\:\:{e}^{\:−{x}} .\mathrm{ln}\:\left(\frac{\:\mathrm{1}}{\:{x}}\:\right)\:{sin}\:\left(\:{x}\:\right)}{{x}\:}\:{dx}\:=\:\frac{\:\pi}{\:\mathrm{8}}\:\left(\:\mathrm{2}\:\gamma\:+\mathrm{ln}\:\left(\mathrm{2}\:\right)\:\right)\:…\blacksquare\:\:\:\:\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:{m}.{n} \\ $$$$ \\ $$…