Question Number 59467 by aliesam last updated on 10/May/19 $$\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{1}}{{cos}\left({x}\right)+{sinh}\left({x}\right)}\:{dx} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 190533 by TUN last updated on 05/Apr/23 Answered by witcher3 last updated on 05/Apr/23 $$\frac{\mathrm{arctan}\left(\mathrm{x}\right)}{\mathrm{x}}=\underset{\mathrm{0}} {\int}^{\mathrm{1}} \frac{\mathrm{da}}{\mathrm{1}+\mathrm{a}^{\mathrm{2}} \mathrm{x}^{\mathrm{2}} } \\ $$$$\Leftrightarrow\Omega=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{tan}^{−\mathrm{1}}…
Question Number 124980 by mathmax by abdo last updated on 07/Dec/20 $$\mathrm{find}\:\int\:\:\frac{\mathrm{dx}}{\left(\sqrt{\mathrm{x}−\mathrm{1}}+\mathrm{2}\sqrt{\mathrm{x}+\mathrm{1}}\right)^{\mathrm{2}} } \\ $$ Answered by liberty last updated on 07/Dec/20 $$\: \\ $$$$\left[\:\:\frac{\sqrt{{x}+\mathrm{1}}−\mathrm{2}\sqrt{{x}+\mathrm{1}}}{{x}−\mathrm{1}−\mathrm{4}\left({x}+\mathrm{1}\right)}\:\right]^{\mathrm{2}}…
Question Number 124985 by Study last updated on 07/Dec/20 Commented by Study last updated on 07/Dec/20 $${help}\:{me} \\ $$ Commented by MJS_new last updated on…
Question Number 124979 by mathmax by abdo last updated on 07/Dec/20 $$\mathrm{let}\:\mathrm{f}\left(\mathrm{x}\right)=\mathrm{arctan}\left(\mathrm{x}^{\mathrm{n}} \right)\:\mathrm{with}\:\mathrm{n}\:\mathrm{natural} \\ $$$$\left.\mathrm{1}\right)\:\mathrm{find}\:\mathrm{f}^{\left(\mathrm{n}\right)} \left(\mathrm{0}\right)\:\mathrm{and}\:\mathrm{f}^{\left(\mathrm{n}\right)} \left(\mathrm{1}\right) \\ $$$$\left.\mathrm{2}\right)\mathrm{developp}\:\mathrm{f}\:\mathrm{at}\:\mathrm{integr}\:\mathrm{serie} \\ $$$$\left.\mathrm{3}\right)\mathrm{calculte}\:\:\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{f}\left(\mathrm{x}\right)}{\mathrm{x}^{\mathrm{n}} }\mathrm{dx}\:\mathrm{with}\:\mathrm{n}\geqslant\mathrm{2} \\…
Question Number 124976 by liberty last updated on 07/Dec/20 $$\left(\mathrm{1}\right)\:{The}\:{gravitational}\:{force}\:\left({in}\:{lb}\right)\:{of} \\ $$$${attraction}\:{between}\:{two}\:{objects}\:{is}\:{given} \\ $$$${by}\:{F}\:=\frac{{k}}{{x}^{\mathrm{2}} },\:{where}\:{x}\:{is}\:{the}\:{distance} \\ $$$${between}\:{the}\:{objects}.\:{If}\:{the}\:{objects}\:{are} \\ $$$$\mathrm{10}\:{ft}\:{apart},\:{find}\:{the}\:{work}\:{required}\:{to} \\ $$$${separate}\:{them}\:{until}\:{they}\:{are}\:\mathrm{50}\:{ft}\:{apart}.\:{Express} \\ $$$${the}\:{result}\:{in}\:{terms}\:{of}\:{k}. \\ $$$$\left({a}\right)\:\frac{{k}}{\mathrm{500}}\:\:\:\:\:\:\left({b}\right)\:\frac{\mathrm{2}{k}}{\mathrm{25}}\:\:\:\:\:\left({c}\right)\:\frac{{k}}{\mathrm{5}}\:\:\:\left({d}\right)\:\frac{{k}}{\mathrm{40}}…
Question Number 59438 by tanmay last updated on 10/May/19 Commented by maxmathsup by imad last updated on 10/May/19 $${we}\:{have}\:{I}_{{n}} =_{\pi{x}\:={t}} \:\:\:\:\int_{\mathrm{0}} ^{\pi} \:\:\frac{{t}^{{n}} }{\pi^{{n}} }\:{sin}\left({t}\right)\:\frac{{dt}}{\pi}\:=\frac{\mathrm{1}}{\pi^{{n}+\mathrm{1}}…
Question Number 124957 by Study last updated on 07/Dec/20 Commented by Study last updated on 07/Dec/20 $${please}\:{help}\:{me}?? \\ $$ Answered by Dwaipayan Shikari last updated…
Question Number 190487 by stvnmaxi last updated on 03/Apr/23 Answered by aleks041103 last updated on 04/Apr/23 $${A}=\int_{\mathrm{0}} ^{\mathrm{1}} \int_{\mathrm{0}} ^{\mathrm{1}} {x}^{\mathrm{2}} {y}^{\mathrm{2}} {dxdy}\:=\left(\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{\mathrm{2}}…
Question Number 124922 by mathmax by abdo last updated on 07/Dec/20 $$\mathrm{let}\:\mathrm{f}\left(\mathrm{x}\right)=\frac{\mathrm{ln}\left(\mathrm{1}+\mathrm{2x}\right)}{\mathrm{x}^{\mathrm{2}} +\mathrm{1}} \\ $$$$\left.\mathrm{1}\right)\:\mathrm{calculste}\:\mathrm{f}^{\left(\mathrm{n}\right)} \left(\mathrm{x}\right)\:\mathrm{and}\:\mathrm{f}^{\left(\mathrm{n}\right)} \left(\mathrm{0}\right) \\ $$$$\left.\mathrm{2}\right)\mathrm{develop}\:\mathrm{f}\:\mathrm{at}\:\mathrm{integr}\:\mathrm{serie} \\ $$$$\left.\mathrm{3}\right)\:\mathrm{find}\:\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{f}\left(\mathrm{x}\right)\mathrm{dx} \\ $$…