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Question Number 184098 by paul2222 last updated on 02/Jan/23

$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\boldsymbol{\mathrm{ln}}^{\mathrm{2}} \left(\mathrm{1}−\boldsymbol{\mathrm{x}}\right)}{\mathrm{1}−\boldsymbol{\mathrm{x}}}\boldsymbol{\mathrm{dx}} \\$$

Commented bypaul2222 last updated on 02/Jan/23

$$\boldsymbol{\mathrm{let}}\:\mathrm{1}−\boldsymbol{\mathrm{x}}=\boldsymbol{\mathrm{u}} \\$$ $$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\boldsymbol{\mathrm{ln}}^{\mathrm{2}} \left(\boldsymbol{\mathrm{u}}\right)}{\boldsymbol{\mathrm{u}}}\boldsymbol{\mathrm{du}} \\$$ $$\boldsymbol{\mathrm{let}}\:\boldsymbol{\mathrm{ln}}\left(\boldsymbol{\mathrm{u}}\right)=−\boldsymbol{\mathrm{k}}\:\boldsymbol{\mathrm{du}}=−\boldsymbol{\mathrm{e}}^{−\boldsymbol{\mathrm{k}}} \boldsymbol{\mathrm{dk}} \\$$ $$\int_{\mathrm{0}} ^{\infty} \frac{\boldsymbol{\mathrm{k}}^{\mathrm{2}} }{\boldsymbol{\mathrm{e}}^{−\boldsymbol{\mathrm{k}}} }\boldsymbol{\mathrm{e}}^{−\boldsymbol{\mathrm{k}}} \boldsymbol{\mathrm{dk}}=\int_{\mathrm{0}} ^{\infty} \boldsymbol{\mathrm{k}}^{\mathrm{2}} \boldsymbol{\mathrm{dk}} \\$$ $$\left(\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{integral}}\:\boldsymbol{\mathrm{diverges}}\right) \\$$

Commented bymokys last updated on 03/Jan/23

$${e}^{{y}} \:=\:\mathrm{1}\:−\:{x}\:\rightarrow\:{e}^{{y}} {dy}\:=\:−{dx} \\$$ $$\\$$ $$\:{x}\:=\:\mathrm{1}\:\rightarrow\:{y}\:=\:−\infty\:\:\:,\:\:{x}\:=\:\mathrm{0}\:\rightarrow\:{y}\:=\:\mathrm{0} \\$$ $$\\$$ $$\:\int_{−\infty} ^{\:\mathrm{0}} \:{y}^{\mathrm{2}} \:{dy}\:{so}\:{is}\:{diverge}\: \\$$

Answered by Ml last updated on 03/Jan/23

$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\left(\mathrm{ln}\left(\mathrm{1}−\mathrm{x}\right)\right)^{\mathrm{2}} }{\left(\mathrm{1}−\mathrm{x}\right)}\mathrm{dx} \\$$ $$\begin{cases}{\mathrm{u}=\mathrm{ln}\left(\mathrm{1}−\mathrm{x}\right)}\\{\mathrm{du}=−\frac{\mathrm{1}}{\left(\mathrm{1}−\mathrm{x}\right)}\mathrm{dx}}\end{cases}\:,\:\mathrm{dx}=−\left(\mathrm{1}−\mathrm{x}\right)\mathrm{du} \\$$ $$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{u}^{\mathrm{2}} \:\left(−\left(\mathrm{1}−\mathrm{x}\right)\right)\mathrm{du}}{\left(\mathrm{1}−\mathrm{x}\right)}=−\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{u}^{\mathrm{2}} \mathrm{du}=−\frac{\mathrm{1}}{\mathrm{3}}\mathrm{u}^{\mathrm{3}} \\$$ $$\Rightarrow\:−\frac{\mathrm{1}}{\mathrm{3}}\left(\mathrm{ln}\left(\mathrm{1}−\mathrm{x}\right)\right)^{\mathrm{3}} \mid_{\mathrm{0}} ^{\mathrm{1}} =\infty \\$$