Category: Integration

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 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 25796 by abdo imad last updated on 15/Dec/17 $$findthevalueoff\left(x\right)={\int }_{0^{}}^{\pi }ln(1+xcos\theta )d\theta with0<x<1$$ Terms of Service Privacy Policy Contact: info@tinkutara.com

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}}} \…

Question Number 156849 by amin96 last updated on 16/Oct/21 $${\int }_0^1\frac{ln(e+\frac{1}{1-t})}{\sqrt{t}}dt=?$$ Terms of Service Privacy Policy Contact: info@tinkutara.com