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integral of ( x^2-2x=1)/x dx
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No boundaries? If not - just find the antiderivative and add a constant C.
x^(3)/3-x^(2)-x+C
Can you show me how you got that?
\[\int\limits (x^2-2 x-1)/x dx = 1/2 ((x-4) x-2 \ln(x))+c\]
Oh, missed the /x, nvm my answer, hehe
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Its x^2-2x+1
You wrote: "x^2-2x=1"
Yes i know i meant to but a +
\[\int x^ndx=\frac{x^{n+1}}{n+1}+C\]
\[\int\limits (x^2-2 x+1)/x dx = 1/2 (x-4) x+\ln(x)+c\]
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