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∫_(-1)^1 {dx/ (3x-4)} is this the ln|x| antiderivative for 1/u?
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u=3x-4 du=3 ∫ 1/u du lnx+c then substitute ln(3x-4)+c then ln (3(1)-4)/3-[ln(3(-1)-4]/3 -0.64
it could be, but your missing a 3 up top
we multiply it by 3/3 and use the top to calibrate for the missing 3; the other 3 can be pulled out as a constant (1/3) to be put back later
okay thank you
integration is an art more than a science; there are no rules that will get you an answer every time like a derivative
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lol kk
but if you can multiply by a useful form of 1; or add a useful form of 0; it goes along way
k
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