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Question Number 163357 by mnjuly1970 last updated on 06/Jan/22

∫_0 ^( 1) ln(3x−3x^( 2) + x^( 3) )= ?

∫01ln(3x−3x2+x3)=?

Answered by Ar Brandon last updated on 06/Jan/22

I=∫_0 ^1 ln(3x−3x^2 +x^3 )dx =∫_0 ^1 lnxdx+∫_0 ^1 ln(x^2 −3x+3)dx =[xlnx−x]_0 ^1 +[xln(x^2 −3x+3)]_0 ^1 −∫_0 ^1 ((2x^2 −3x)/(x^2 −3x+3))dx =−1−∫_0 ^1 (2+((3x−6)/(x^2 −3x+3)))dx=−1−2−∫_0 ^1 ((3x−6)/(x^2 −3x+3))dx =−3−(3/2)∫_0 ^1 (((2x−3)/(x^2 −3x+3))−(1/(x^2 −3x+3)))dx =−3−(3/2)[ln(x^2 −3x+3)]_0 ^1 +(3/2)∫_0 ^1 (dx/((x−(3/2))^2 +(3/4))) =−3+(3/2)ln3+(√3)[arctan(((2x−3)/( (√3))))]_0 ^1 =−3+(3/2)ln3+(√3)(−(π/6)+(π/3))=−3+(3/2)ln3+(π/(2(√3)))

I=∫01ln(3x−3x2+x3)dx=∫01lnxdx+∫01ln(x2−3x+3)dx=[xlnx−x]01+[xln(x2−3x+3)]01−∫012x2−3xx2−3x+3dx=−1−∫01(2+3x−6x2−3x+3)dx=−1−2−∫013x−6x2−3x+3dx=−3−32∫01(2x−3x2−3x+3−1x2−3x+3)dx=−3−32[ln(x2−3x+3)]01+32∫01dx(x−32)2+34=−3+32ln3+3[arctan(2x−33)]01=−3+32ln3+3(−π6+π3)=−3+32ln3+π23

Commented by mnjuly1970 last updated on 06/Jan/22

mercey mr brandon..

merceymrbrandon..

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