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Question Number 29162 by abdo imad last updated on 04/Feb/18

find find I= ∫_1 ^3 ((∣x−2∣)/((x^2 −4x)^2 ))dx .

findfindI=∫13∣x−2∣(x2−4x)2dx.

Commented by abdo imad last updated on 06/Feb/18

the chasles relation give I=∫_1 ^2 ((−x+2)/((x^2 −4x)^2 ))dx +∫_2 ^3 ((x−2)/((x^2 −4x)^2 ))dx ∫_1 ^2 ((−x +2)/((x^2 −4x)^2 ))dx=−(1/2)∫_1 ^2 ((2x−4)/((x^2 −4x)^2 ))dx =(1/2)[ (1/(x^2 −4x))]_1 ^2 =(1/2)(−(1/4) +(1/3))=(1/(24)) and ∫_2 ^3 ((x−2)/((x^2 −4x)^2 ))dx=(1/2) ∫_1 ^3 ((2x−4)/((x^2 −4x)^2 ))dx =−(1/2)[ (1/(x^2 −4x))]_1 ^3 =−(1/2)( −(1/3) +(1/3))=0 ⇒ I=(1/(24)) .

thechaslesrelationgiveI=∫12−x+2(x2−4x)2dx+∫23x−2(x2−4x)2dx∫12−x+2(x2−4x)2dx=−12∫122x−4(x2−4x)2dx=12[1x2−4x]12=12(−14+13)=124and∫23x−2(x2−4x)2dx=12∫132x−4(x2−4x)2dx=−12[1x2−4x]13=−12(−13+13)=0⇒I=124.

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