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yeh

for x<0 , |x| = -x

therefore f(x) = ln(x^2) / -x


f(x) = - ln (x^2) / x

u= ln(x^2)
du/dx= 2/x
v= x
dv/dx= 1

f ' (x) = - [ ( 2 - ln(x^2) ) ] / ( x^2 )
sub x=-e into that
 
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similarly , between the limits of -1 and -e , |x| = -x

therefore

we want to integrate - [ln(x^2) / x] dx

then sub u= ln(x^2)

du/dx= 2/x

so dx= x du/ 2

so we want to integrate -u/2 du , and remember to change the limits of the integral
 
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