Question Number 25851 by abdo imad last updated on 15/Dec/17 $${find}\:{the}\:{value}\:{of}\:{integral}\:\:\:\int_{{R}} \left({z}−{a}\right)^{−\mathrm{1}} {dz}\:\:{with}\:{a}\:{from}\:{C}\:\:{aplly} \\ $$$${this}\:{result}\:{to}\:{find}\:{the}\:{value}\:{of}\:\:\:\int_{\mathrm{0}} ^{\infty} \left(\mathrm{2}\:+{x}^{\mathrm{4}_{} } \right)^{−\mathrm{1}} {dx}. \\ $$ Commented by abdo…
Question Number 156914 by cortano last updated on 17/Oct/21 Answered by puissant last updated on 17/Oct/21 $${D}=\int\frac{{arcsin}\left(\frac{\mathrm{1}}{{x}}\right)}{{x}^{\mathrm{5}} }{dx}\:;\:{u}=\frac{\mathrm{1}}{{x}}\:\rightarrow\:{du}=−\frac{\mathrm{1}}{{x}^{\mathrm{2}} }{dx} \\ $$$$\Rightarrow\:{D}=−\int\:\frac{{u}^{\mathrm{5}} {arcsin}\left({u}\right)}{{u}^{\mathrm{2}} }{du}=−\int{u}^{\mathrm{3}} {arcsin}\left({u}\right){du} \\…
Question Number 156915 by mnjuly1970 last updated on 17/Oct/21 Answered by qaz last updated on 07/Nov/21 $$\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{dx}}{\left(\pi^{\mathrm{2}} +\mathrm{x}^{\mathrm{2}} \right)\mathrm{cosh}\:\mathrm{x}} \\ $$$$=\int_{−\infty} ^{+\infty} \frac{\mathrm{e}^{\mathrm{x}}…
Question Number 25837 by mubeen897@hotmail.com last updated on 15/Dec/17 $$\int\left({x}^{{n}} {lnx}\right){dx} \\ $$ Answered by kaivan.ahmadi last updated on 15/Dec/17 $$\mathrm{u}=\mathrm{lnx}\Rightarrow\mathrm{du}=\frac{\mathrm{dx}}{\mathrm{x}} \\ $$$$\mathrm{dv}=\mathrm{x}^{\mathrm{n}} \mathrm{dx}\Rightarrow\mathrm{v}=\frac{\mathrm{x}^{\mathrm{n}+\mathrm{1}} }{\mathrm{n}+\mathrm{1}}…
Question Number 25824 by abdo imad last updated on 15/Dec/17 $${if}\:\:\int_{{R}} \:{e}^{−{x}^{\mathrm{2}} } {dx}\:\:=\:\:\pi^{\mathrm{1}/\mathrm{2}} \:\:\:\:{find}\:{value}\:\:{of}\:\:\:\int_{{R}} \:{a}^{−{x}^{\mathrm{2}_{} } } {dx}\:\:\:{with}\:{a}>\mathrm{0}\:{and}\:{a}\:{not}\:\mathrm{1}. \\ $$ Terms of Service Privacy…
Question Number 25821 by abdo imad last updated on 15/Dec/17 $${answer}\:{to}\:{q}\mathrm{25796}\:{f}_{} {ind}\:{the}\:{value}\:{off}\left({x}\right)=\:\int_{\mathrm{0}^{} } ^{\pi_{} } {ln}\left(\mathrm{1}+{xcos}\theta\right){d}\theta\:{with}\:\mathrm{0}<{x}<\mathrm{1}\:\:\:\partial{f}/\partial{x}=\:\int_{\mathrm{0}} ^{\pi} \:\:{cos}\theta\left(\mathrm{1}+{xcos}\theta\right)^{−\mathrm{1}} {d}\theta=\pi{x}^{−\mathrm{1}} −{x}^{−\mathrm{1}} \int_{\mathrm{0}} ^{\pi} \left(\mathrm{1}+{xcos}\theta\right)^{−\mathrm{1}} {d}\theta{and}\:{by}\:{the}\:{changeent}\:{tan}\theta={u}\:{then}\:{the}\:{changement}\:{u}=\left(\left(\mathrm{1}+{x}\right)\left(\mathrm{1}+{x}\right)^{−\mathrm{1}^{} }…
Question Number 25811 by ajfour last updated on 15/Dec/17 Commented by ajfour last updated on 15/Dec/17 $${Q}.\mathrm{25744}\:\:\:\left({in}\:{response}\:{to}\right) \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 25796 by abdo imad last updated on 15/Dec/17 $${find}\:{the}\:{value}\:{of}\:{f}\left({x}\right)=\int_{\mathrm{0}^{} } ^{\pi} {ln}\left(\:\mathrm{1}+{xcos}\theta\right){d}\theta\:{with}\:\mathrm{0}<{x}<\mathrm{1} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 156869 by cortano last updated on 16/Oct/21 Answered by gsk2684 last updated on 16/Oct/21 $$\int\frac{\mathrm{sec}\:{x}\left(\mathrm{sec}\:{x}\left(\mathrm{sec}\:{x}+\mathrm{tan}\:{x}\right)\right)}{\left(\mathrm{sec}\:{x}+\mathrm{tan}\:{x}\right)^{\mathrm{6}} }{dx} \\ $$$${put}\:\mathrm{sec}\:{x}+\mathrm{tan}\:{x}={t},\mathrm{sec}\:{x}−\mathrm{tan}\:{x}=\frac{\mathrm{1}}{{t}}\Rightarrow \\ $$$$\mathrm{sec}\:{x}\mathrm{tan}\:{x}+\mathrm{sec}\:^{\mathrm{2}} {x}=\frac{{dt}}{{dx}},\:\mathrm{2sec}\:{x}={t}+\frac{\mathrm{1}}{{t}} \\ $$$$\int\frac{{t}+\frac{\mathrm{1}}{{t}}}{\mathrm{2}{t}^{\mathrm{6}}…
Question Number 91326 by mhmd last updated on 29/Apr/20 Commented by mathmax by abdo last updated on 30/Apr/20 $${I}\:=\int\:\:\frac{{dx}}{\left({x}−\mathrm{1}\right)^{\mathrm{2}} \sqrt{{x}^{\mathrm{2}} −{x}+\mathrm{1}}}\:\Rightarrow\:{I}\:=_{{x}−\mathrm{1}={t}} \:\:\:\:\int\:\:\frac{{dt}}{{t}^{\mathrm{2}} \sqrt{\left({t}+\mathrm{1}\right)^{\mathrm{2}} −\left({t}+\mathrm{1}\right)+\mathrm{1}}} \\…