Question Number 943 by miguel last updated on 04/May/15
$${help}\:{me}.. \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\int\frac{{lnx}}{\:\sqrt{\mathrm{1}−{x}}}{dx}=??? \\ $$$$ \\ $$
Commented by 123456 last updated on 04/May/15
$$\mathrm{0}<{x}<\mathrm{1}\:\:\:? \\ $$$${u}=\mathrm{ln}\:{x}\Leftrightarrow{x}={e}^{{u}} \\ $$$${du}=\frac{{dx}}{{x}}=\frac{{dx}}{{e}^{{u}} } \\ $$$$\int\frac{\mathrm{ln}\:{x}}{\:\sqrt{\mathrm{1}−{x}}}{dx}\overset{?} {=}\int\frac{{ue}^{{u}} }{\:\sqrt{\mathrm{1}−{e}^{{u}} }}{du} \\ $$
Answered by prakash jain last updated on 04/May/15
![Integrat by parts −2ln x(√(1−x))+2∫ ((√(1−x))/x) dx ∫((√(1−x))/x) dx=∫ ((1−x)/(x(√(1−x))))dx =−∫(1/( (√(1−x)))) dx+∫(1/(x(√(1−x))))dx =−2(√(1−x))+∫ (1/(x(√(1−x))))dx ∫ (1/(x(√(1−x))))dx, put 1−x=u^2 , x=1−u^2 , dx=−2udu ∫ ((−2udu)/((1−u^2 )u))=−∫((2du)/((1−u^2 )))=−[∫(du/(1−u))+∫(du/(1+u))] =ln (1−u)−ln (1+u)=ln (1−(√(1−x)))−ln (1+(√(1+x)))](https://www.tinkutara.com/question/Q946.png)
$$\mathrm{Integrat}\:\mathrm{by}\:\mathrm{parts}\: \\ $$$$−\mathrm{2ln}\:{x}\sqrt{\mathrm{1}−{x}}+\mathrm{2}\int\:\frac{\sqrt{\mathrm{1}−{x}}}{{x}}\:{dx} \\ $$$$\int\frac{\sqrt{\mathrm{1}−{x}}}{{x}}\:{dx}=\int\:\frac{\mathrm{1}−{x}}{{x}\sqrt{\mathrm{1}−{x}}}{dx} \\ $$$$=−\int\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}−{x}}}\:{dx}+\int\frac{\mathrm{1}}{{x}\sqrt{\mathrm{1}−{x}}}{dx} \\ $$$$=−\mathrm{2}\sqrt{\mathrm{1}−{x}}+\int\:\frac{\mathrm{1}}{{x}\sqrt{\mathrm{1}−{x}}}{dx} \\ $$$$\int\:\frac{\mathrm{1}}{{x}\sqrt{\mathrm{1}−{x}}}{dx},\:\mathrm{put}\:\mathrm{1}−{x}={u}^{\mathrm{2}} ,\:{x}=\mathrm{1}−{u}^{\mathrm{2}} ,\:{dx}=−\mathrm{2}{udu} \\ $$$$\int\:\frac{−\mathrm{2}{udu}}{\left(\mathrm{1}−{u}^{\mathrm{2}} \right){u}}=−\int\frac{\mathrm{2}{du}}{\left(\mathrm{1}−{u}^{\mathrm{2}} \right)}=−\left[\int\frac{{du}}{\mathrm{1}−{u}}+\int\frac{{du}}{\mathrm{1}+{u}}\right] \\ $$$$=\mathrm{ln}\:\left(\mathrm{1}−{u}\right)−\mathrm{ln}\:\left(\mathrm{1}+{u}\right)=\mathrm{ln}\:\left(\mathrm{1}−\sqrt{\mathrm{1}−{x}}\right)−\mathrm{ln}\:\left(\mathrm{1}+\sqrt{\mathrm{1}+{x}}\right) \\ $$
Commented by prakash jain last updated on 04/May/15
$$\mathrm{ln}\:{x}\:\mathrm{is}\:\mathrm{first}\:\mathrm{funcion}\:\mathrm{and}\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}−{x}}}\:\:\mathrm{is}\:\mathrm{the}\:\mathrm{second} \\ $$$$\mathrm{function}\:\mathrm{for}\:\mathrm{integrating}\:\mathrm{by}\:\mathrm{parts}. \\ $$