Question Number 67921 by mhmd last updated on 02/Sep/19 $$\int{e}^{{y}^{\mathrm{2}} /\mathrm{2}} {dy} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 133452 by bagjagunawan last updated on 22/Feb/21 Answered by mnjuly1970 last updated on 22/Feb/21 $$\boldsymbol{\phi}\overset{{x}=\mathrm{2}{y}−\mathrm{1}} {=}\int_{\frac{\mathrm{1}}{\mathrm{2}}\:} ^{\:\mathrm{1}} \frac{{ln}\left(\mathrm{2}−\mathrm{2}{y}\right){ln}\left({ln}\left(\mathrm{2}{y}\right)\right)}{\mathrm{2}{y}}\left(\mathrm{2}\right){dy} \\ $$$$=\int_{\frac{\mathrm{1}}{\mathrm{2}}} ^{\:\mathrm{1}} \left({ln}\left(\mathrm{2}\right)+{ln}\left(\mathrm{1}−{y}\right)\right)\left({ln}\left(\mathrm{2}\right)+{ln}\left({y}\right)\right)\frac{{dy}}{{y}} \\…
Question Number 67919 by aliesam last updated on 02/Sep/19 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 67907 by A8;15: last updated on 02/Sep/19 Commented by mathmax by abdo last updated on 02/Sep/19 $${let}\:{I}\:=\int\:\frac{{dx}}{{x}\sqrt{{x}^{\mathrm{2}} +{x}−\mathrm{6}}} \\ $$$${x}^{\mathrm{2}} +{x}−\mathrm{6}=\mathrm{0}\rightarrow\Delta=\mathrm{1}−\mathrm{4}\left(−\mathrm{6}\right)\:=\mathrm{25}\:\Rightarrow{x}_{\mathrm{1}} =\frac{−\mathrm{1}+\mathrm{5}}{\mathrm{2}}=\mathrm{2}\:\:{and}\: \\…
Question Number 133440 by bagjagunawan last updated on 22/Feb/21 Answered by Ar Brandon last updated on 22/Feb/21 $$\mathrm{f}\left({x}\right)={x}+\mathrm{5}+\sqrt{\mathrm{8}{x}}+\sqrt{\mathrm{12}{x}}+\sqrt{\mathrm{24}}\: \\ $$$$\:\:\:\:\:\:\:\:\:={x}+\left(\sqrt{\mathrm{8}}+\sqrt{\mathrm{12}}\right)\sqrt{{x}}+\mathrm{5}+\sqrt{\mathrm{24}}=\left(\sqrt{{x}}+\sqrt{\mathrm{2}}+\sqrt{\mathrm{3}}\right)^{\mathrm{2}} \\ $$$$\mathcal{I}=\int\mathrm{f}\left({x}\right)\mathrm{d}{x}=\int\sqrt{\left(\sqrt{{x}}+\sqrt{\mathrm{2}}+\sqrt{\mathrm{3}}\right)^{\mathrm{2}} }\mathrm{d}{x} \\ $$$$\:\:\:=\int\left(\sqrt{{x}}+\sqrt{\mathrm{2}}+\sqrt{\mathrm{3}}\right)\mathrm{d}{x}=\frac{\mathrm{2}}{\mathrm{3}}\sqrt{{x}^{\mathrm{3}}…
Question Number 133433 by greg_ed last updated on 22/Feb/21 $$\boldsymbol{\mathrm{hi}},\:\boldsymbol{\mathrm{everybody}}\:! \\ $$$$\boldsymbol{\mathrm{with}}\:\boldsymbol{\mathrm{I}}=\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{1}}{\boldsymbol{{x}}^{\mathrm{4}} +\mathrm{1}}\:\boldsymbol{{dx}}, \\ $$$$\boldsymbol{\mathrm{prove}}\:\boldsymbol{\mathrm{that}}\::\:\mathrm{2}\boldsymbol{\mathrm{I}}=\int_{\mathrm{0}} ^{\infty} \:\frac{\boldsymbol{{x}}^{\mathrm{2}} +\mathrm{1}}{\boldsymbol{{x}}^{\mathrm{4}} +\mathrm{1}}\:\boldsymbol{{dx}}. \\ $$ Answered by…
Question Number 133426 by metamorfose last updated on 22/Feb/21 $$\int\lfloor{x}\rfloor{dx}=?… \\ $$ Answered by MJS_new last updated on 22/Feb/21 $$\mathrm{for}\:{a}<{b}:\:\underset{{a}} {\overset{{b}} {\int}}\lfloor{x}\rfloor{dx}=\lfloor{a}\rfloor\underset{{a}} {\overset{\lceil{a}\rceil} {\int}}{dx}+\underset{\lceil{a}\rceil} {\overset{\lfloor{b}\rfloor−\mathrm{1}}…
Question Number 2340 by Yozzi last updated on 18/Nov/15 $${Prove}\:{that},\:\forall{m}\in\mathbb{Z}^{+} , \\ $$$$\underset{{r}=\mathrm{1}} {\overset{{m}} {\prod}}\Gamma\left({x}+\frac{{r}−\mathrm{1}}{{m}}\right)={m}^{\frac{\mathrm{1}}{\mathrm{2}}−{mx}} \left(\mathrm{2}\pi\right)^{\frac{{m}−\mathrm{1}}{\mathrm{2}}} \Gamma\left({mx}\right). \\ $$$$\left\{\Gamma\left({x}\right)=\int_{\mathrm{0}} ^{\infty} {t}^{{x}−\mathrm{1}} {e}^{−{t}} {dt},\:{x}>\mathrm{0}\right\} \\ $$…
Question Number 67851 by mathmax by abdo last updated on 01/Sep/19 $${find}\:\int\:\:\frac{{dx}}{{x}^{\mathrm{2}} −{z}}\:\:{with}\:{z}\:{from}\:{C}\:. \\ $$ Commented by MJS last updated on 01/Sep/19 $$\mathrm{what}'\mathrm{s}\:\mathrm{the}\:\mathrm{problem}/\mathrm{mistake}\:\mathrm{if}\:\mathrm{we}\:\mathrm{just} \\ $$$$\mathrm{calculate}\:\mathrm{it}\:\mathrm{as}\:\mathrm{if}\:{z}\in\mathbb{R}?…
Question Number 67850 by mathmax by abdo last updated on 01/Sep/19 $${calculate}\:\int_{−\infty} ^{+\infty} \:\:\frac{{dx}}{{x}^{\mathrm{2}} −{z}}\:\:{with}\:{z}\:{from}\:{C} \\ $$ Commented by mathmax by abdo last updated on…