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Question Number 75141 by chess1 last updated on 07/Dec/19

Commented by chess1 last updated on 08/Dec/19

solve please

solveplease

Commented by chess1 last updated on 08/Dec/19

Sir mathmax , sir mind is power

Sirmathmax,sirmindispower

Answered by mind is power last updated on 08/Dec/19

∫ln(z−a)dz=(z−a)ln(z−a)−z ⇒∫_e ^(x+y+e) ((ln(z−x−y))/((x−e)(x+y−e)))dz=(([(z−x−y)ln(z−x−y)−z]_e ^(x+y+e) )/((x−e)(x+y−e))) =((e−(x+y+e)−(e−x−y)ln(e−x−y)+e)/((x−e)(e−x−y))) =((1−ln(e−x−y))/(x−e)) ∫_0 ^(e−x−1) ((1−ln(e−x−y))/(x−e))dy=(1/(x−e))∫_0 ^(e−x−1) (1−ln(e−x−y))dy =(1/(x−e))[y+(e−x−y)ln(e−x−y)+y]_0 ^(e−x−1) =(1/(x−e))[2(e−x−1)−(e−x)ln(e−x)] =2−(2/(x−e))−ln(e−x) ∫_0 ^(e−1) (2−(2/(x−e))−ln(e−x))dx=[2x−2ln∣x−e∣+(e−x)ln(e−x)+x]_0 ^(e−1) =3(e−1)+2−e=2e−1

∫ln(z−a)dz=(z−a)ln(z−a)−z⇒∫ex+y+eln(z−x−y)(x−e)(x+y−e)dz=[(z−x−y)ln(z−x−y)−z]ex+y+e(x−e)(x+y−e)=e−(x+y+e)−(e−x−y)ln(e−x−y)+e(x−e)(e−x−y)=1−ln(e−x−y)x−e∫0e−x−11−ln(e−x−y)x−edy=1x−e∫0e−x−1(1−ln(e−x−y))dy=1x−e[y+(e−x−y)ln(e−x−y)+y]0e−x−1=1x−e[2(e−x−1)−(e−x)ln(e−x)]=2−2x−e−ln(e−x)∫0e−1(2−2x−e−ln(e−x))dx=[2x−2ln∣x−e∣+(e−x)ln(e−x)+x]0e−1=3(e−1)+2−e=2e−1

Commented by chess1 last updated on 08/Dec/19

great!

great!

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