Question Number 63410 by aliesam last updated on 03/Jul/19 Commented by mathmax by abdo last updated on 03/Jul/19 $$\left.\mathrm{2}\right)\:{let}\:{I}\:=\int\:\:\:\frac{{xdx}}{\mathrm{1}+{sinx}}\:\:\:{changement}\:{tan}\left(\frac{{x}}{\mathrm{2}}\right)={t}\:{give}\: \\ $$$${I}\:=\:\int\:\:\:\frac{\mathrm{2}{arctan}\left({t}\right)}{\left(\mathrm{1}+\frac{\mathrm{2}{t}}{\mathrm{1}+{t}^{\mathrm{2}} }\right)\left(\mathrm{1}+{t}^{\mathrm{2}} \right)}{dt}\:=\:\mathrm{2}\int\:\:\:\:\frac{{arctan}\left({t}\right)}{\mathrm{1}+{t}^{\mathrm{2}} \:+\mathrm{2}{t}}\:{dt}\:=\mathrm{2}\:\int\:\:\:\frac{{arctan}\left({t}\right)}{\left({t}+\mathrm{1}\right)^{\mathrm{2}} }\:{dt}\:\:…
Question Number 128943 by bramlexs22 last updated on 11/Jan/21 $$\:\int\:\frac{\mathrm{x}^{\mathrm{9}} }{\left(\mathrm{x}^{\mathrm{2}} +\mathrm{m}^{\mathrm{2}} \right)\mathrm{sec}\:\left(\mathrm{5x}\right)}\:\mathrm{dx}\:? \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 63405 by mathmax by abdo last updated on 03/Jul/19 $${find}\:\int\:\sqrt{\frac{{x}^{\mathrm{2}} −\mathrm{4}{x}+\mathrm{1}}{{x}+\mathrm{2}}}{dx} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 63404 by mathmax by abdo last updated on 03/Jul/19 $$\left.\mathrm{1}\right)\:{find}\:\:\int\:\:\:\:\frac{{x}+\mathrm{1}}{{x}^{\mathrm{3}} −\mathrm{3}{x}\:−\mathrm{2}}{dx} \\ $$$$\left.\mathrm{2}\right)\:{calculate}\:\:\int_{\mathrm{4}} ^{+\infty} \:\:\:\:\:\frac{{x}+\mathrm{1}}{{x}^{\mathrm{3}} −\mathrm{3}{x}\:+\mathrm{2}}{dx} \\ $$ Commented by MJS last updated…
Question Number 63395 by mathmax by abdo last updated on 03/Jul/19 $${let}\:{f}\left({t}\right)\:=\int_{\mathrm{0}} ^{\infty} \:\frac{{ln}\left(\mathrm{1}+{tx}\right)}{\mathrm{1}+{x}^{\mathrm{2}} }{dx}\:\:\:{with}\:\:\mid{t}\mid<\mathrm{1} \\ $$$$\left.\mathrm{1}\right)\:{determine}\:{a}\:{explicit}\:\:{form}\:{of}\:{f}\left({t}\right) \\ $$$$\left.\mathrm{2}\right)\:{find}\:{the}\:{value}\:{of}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{ln}\left(\mathrm{1}+{x}\right)}{\mathrm{1}+{x}^{\mathrm{2}} }{dx} \\ $$ Commented…
Question Number 128926 by liberty last updated on 11/Jan/21 $$\:\beta\:=\:\int_{\mathrm{0}} ^{\:\mathrm{5}} \:\frac{{dx}}{{x}^{\mathrm{3}} \left(\mathrm{1}+{e}^{{x}} \right)^{\mathrm{2}} }\:? \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 128922 by mnjuly1970 last updated on 11/Jan/21 $$\:\:\:\:\:\:\:\:\:\:\:\:\:…{advanced}\:\:{calculus}… \\ $$$$\:\:\:\:\:{evaluate}\::::\: \\ $$$$\:\:\:\:\:\:\:\Omega=\int_{\mathrm{0}\:} ^{\:\frac{\mathrm{1}}{\mathrm{2}}} \frac{\mathrm{Arctanh}\left({x}\right)}{{x}}{dx}=? \\ $$$$ \\ $$ Answered by Dwaipayan Shikari last…
Question Number 63372 by necx1 last updated on 03/Jul/19 $${For}\:{what}\:{values}\:{of}\:{a}\:{and}\:{b}\:{will}\:{the} \\ $$$${integral}\:\int_{{a}} ^{{b}} \sqrt{\mathrm{10}−{x}−{x}^{\mathrm{2}} }{dx}\:{be}\:{at} \\ $$$${maximum} \\ $$ Commented by mr W last updated…
Question Number 128900 by bemath last updated on 11/Jan/21 $$\:\int\:\frac{\mathrm{4x}+\mathrm{5}}{\left(\mathrm{x}+\mathrm{2}\right)\left(\mathrm{x}+\mathrm{3}\right)\left(\mathrm{x}+\mathrm{4}\right)\left(\mathrm{x}+\mathrm{5}\right)+\mathrm{1}}\:\mathrm{dx}?\: \\ $$ Answered by liberty last updated on 11/Jan/21 $$\:\int\:\frac{\mathrm{4x}+\mathrm{5}}{\left(\mathrm{x}^{\mathrm{2}} +\mathrm{7x}+\mathrm{10}\right)\left(\mathrm{x}^{\mathrm{2}} +\mathrm{7x}+\mathrm{12}\right)+\mathrm{1}}\:\mathrm{dx}=\left(\ast\right) \\ $$$$\:\left(\mathrm{x}^{\mathrm{2}} +\mathrm{7x}+\mathrm{10}\right)\left(\mathrm{x}^{\mathrm{2}}…
Question Number 128888 by bemath last updated on 11/Jan/21 $$\:\int\:\frac{\left(\mathrm{x}−\mathrm{1}\right)\left(\mathrm{x}−\mathrm{2}\right)\left(\mathrm{x}−\mathrm{3}\right)}{\left(\mathrm{x}−\mathrm{4}\right)\left(\mathrm{x}−\mathrm{5}\right)\left(\mathrm{x}−\mathrm{6}\right)}\:\mathrm{dx}\:=? \\ $$ Answered by Olaf last updated on 11/Jan/21 $$\Omega\:=\:\int\frac{\left({x}−\mathrm{1}\right)\left({x}−\mathrm{2}\right)\left({x}−\mathrm{3}\right)}{\left({x}−\mathrm{4}\right)\left({x}−\mathrm{5}\right)\left({x}−\mathrm{6}\right)}{dx} \\ $$$$\Omega\:=\:\int\left(\mathrm{1}+\frac{\mathrm{A}}{{x}−\mathrm{4}}+\frac{\mathrm{B}}{{x}−\mathrm{5}}+\frac{\mathrm{C}}{{x}−\mathrm{6}}\right){dx} \\ $$$$\mathrm{A}\:=\:\frac{\mathrm{3}×\mathrm{2}×\mathrm{1}}{\left(−\mathrm{1}\right)\left(−\mathrm{2}\right)}\:=\:\mathrm{3} \\…