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Question Number 169863 by SANOGO last updated on 11/May/22

∫_o ^1 xln∣x^2 −2x∣dx

∫o1xln∣x2−2x∣dx

Answered by Mathspace last updated on 11/May/22

x^2 −2x=x(x−2)<0 sur [01] ⇒ ∫_0 ^1 xln∣x^2 −2x∣dx= =∫_0 ^1 xln(2x−x^2 )dx =[(x^2 /2)ln(2x−x^2 )]_0 ^1 −∫_0 ^1 (x^2 /2)×((2−2x)/(2x−x^2 ))dx =o−∫_0 ^1 ((x^2 (1−x))/(2x−x^2 ))dx =−∫_0 ^1 ((x(1−x))/(2−x))dx =−∫_0 ^1 ((x−x^2 )/(2−x))dx=−∫_0 ^1 ((x^2 −x)/(x−2))dx =−∫_0 ^1 ((x^2 −4+4)/(x−2))dx =−∫_0 ^1 (x+2)dx−4∫_0 ^1 (dx/(x−2)) =−[ (x^2 /2)+2x]_0 ^1 −4[ln∣x−2∣]_0 ^1 =−((1/2)+2)−4(−ln(2)) =−(5/2)+4ln(2)

x2−2x=x(x−2)<0sur[01]⇒∫01xln∣x2−2x∣dx==∫01xln(2x−x2)dx=[x22ln(2x−x2)]01−∫01x22×2−2x2x−x2dx=o−∫01x2(1−x)2x−x2dx=−∫01x(1−x)2−xdx=−∫01x−x22−xdx=−∫01x2−xx−2dx=−∫01x2−4+4x−2dx=−∫01(x+2)dx−4∫01dxx−2=−[x22+2x]01−4[ln∣x−2∣]01=−(12+2)−4(−ln(2))=−52+4ln(2)

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