une primitive de ln(1−x^2 )dx puis la convergence de ∫


Question and Answers Forum

All Questions Topic List
None Questions
Previous in All Question Next in All Question
Previous in None Next in None
Question Number 164471 by SANOGO last updated on 17/Jan/22

$u n e p r i m i t i v e d e l n (1 - x^{2}) d x$ $p u i s l a c o n v e r g e n c e d e \int_{0}^{1} l n (1 - x^{2}) d x$
	Answered by Ar Brandon last updated on 18/Jan/22
	$I=∫ln(1−x^2 )dx { ((u(x)=ln(1−x^2 ))),((v′(x)=1)) :} ⇒ { ((u′(x)=((2x)/(x^2 −1)))),((v(x)=x)) :} I=xln(1−x^2 )−2∫(x^2 /(x^2 −1))dx =xln(1−x^2 )−2∫(1+(1/(x^2 −1)))dx =xln(1−x^2 )−∫(2+(1/(x−1))−(1/(x+1)))dx =xln(1−x^2 )−2x+ln∣((x+1)/(x−1))∣$
	$I = \int \ln (1 - x^{2}) d x$ ${\begin{cases} u (x) = \ln (1 - x^{2}) \\ v^{'} (x) = 1 \end{cases} \Rightarrow {\begin{cases} u^{'} (x) = \frac{2 x}{x^{2} - 1} \\ v (x) = x \end{cases}$ $I = x \ln (1 - x^{2}) - 2 \int \frac{x^{2}}{x^{2} - 1} d x$ $= x \ln (1 - x^{2}) - 2 \int (1 + \frac{1}{x^{2} - 1}) d x$ $= x \ln (1 - x^{2}) - \int (2 + \frac{1}{x - 1} - \frac{1}{x + 1}) d x$ $= x \ln (1 - x^{2}) - 2 x + \ln ∣ \frac{x + 1}{x - 1} ∣$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum