help-me-lnx-1-x-dx-

Question Number 943 by miguel last updated on 04/May/15

h e l p m e . .

\int \frac{l n x}{\sqrt{1 - x}} d x = ? ? ?

Commented by 123456 last updated on 04/May/15

0 < x < 1 ?

u = \ln x \Leftrightarrow x = e^{u}

d u = \frac{d x}{x} = \frac{d x}{e^{u}}

\int \frac{\ln x}{\sqrt{1 - x}} d x \overset{?}{=} \int \frac{{u e}^{u}}{\sqrt{1 - e^{u}}} d u

Answered by prakash jain last updated on 04/May/15

Integrat by parts

- 2 \ln x \sqrt{1 - x} + 2 \int \frac{\sqrt{1 - x}}{x} d x

\int \frac{\sqrt{1 - x}}{x} d x = \int \frac{1 - x}{x \sqrt{1 - x}} d x

= - \int \frac{1}{\sqrt{1 - x}} d x + \int \frac{1}{x \sqrt{1 - x}} d x

= - 2 \sqrt{1 - x} + \int \frac{1}{x \sqrt{1 - x}} d x

\int \frac{1}{x \sqrt{1 - x}} d x, put 1 - x = u^{2}, x = 1 - u^{2}, d x = - 2 u d u

\int \frac{- 2 u d u}{(1 - u^{2}) u} = - \int \frac{2 d u}{(1 - u^{2})} = - [\int \frac{d u}{1 - u} + \int \frac{d u}{1 + u}]

= \ln (1 - u) - \ln (1 + u) = \ln (1 - \sqrt{1 - x}) - \ln (1 + \sqrt{1 + x})

Commented by prakash jain last updated on 04/May/15

\ln x is first funcion and \frac{1}{\sqrt{1 - x}} is the second

function for integrating by parts .