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Question Number 164471 by SANOGO last updated on 17/Jan/22

une primitive de ln(1−x^2 )dx puis la convergence de ∫_0 ^1 ln(1−x^2 )dx

$$\:{une}\:{primitive}\:{de}\:{ln}\left(\mathrm{1}−{x}^{\mathrm{2}} \right){dx} \\ $$$${puis}\:{la}\:{convergence}\:{de}\:\int_{\mathrm{0}} ^{\mathrm{1}} {ln}\left(\mathrm{1}−{x}^{\mathrm{2}} \right){dx} \\ $$

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))∣

$${I}=\int\mathrm{ln}\left(\mathrm{1}−{x}^{\mathrm{2}} \right){dx} \\ $$$$\begin{cases}{\mathrm{u}\left({x}\right)=\mathrm{ln}\left(\mathrm{1}−{x}^{\mathrm{2}} \right)}\\{{v}'\left({x}\right)=\mathrm{1}}\end{cases}\:\Rightarrow\begin{cases}{{u}'\left({x}\right)=\frac{\mathrm{2}{x}}{{x}^{\mathrm{2}} −\mathrm{1}}}\\{{v}\left({x}\right)={x}}\end{cases} \\ $$$${I}={x}\mathrm{ln}\left(\mathrm{1}−{x}^{\mathrm{2}} \right)−\mathrm{2}\int\frac{{x}^{\mathrm{2}} }{{x}^{\mathrm{2}} −\mathrm{1}}{dx} \\ $$$$\:\:\:={x}\mathrm{ln}\left(\mathrm{1}−{x}^{\mathrm{2}} \right)−\mathrm{2}\int\left(\mathrm{1}+\frac{\mathrm{1}}{{x}^{\mathrm{2}} −\mathrm{1}}\right){dx} \\ $$$$\:\:\:={x}\mathrm{ln}\left(\mathrm{1}−{x}^{\mathrm{2}} \right)−\int\left(\mathrm{2}+\frac{\mathrm{1}}{{x}−\mathrm{1}}−\frac{\mathrm{1}}{{x}+\mathrm{1}}\right){dx} \\ $$$$\:\:\:={x}\mathrm{ln}\left(\mathrm{1}−{x}^{\mathrm{2}} \right)−\mathrm{2}{x}+\mathrm{ln}\mid\frac{{x}+\mathrm{1}}{{x}−\mathrm{1}}\mid \\ $$

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