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How to integrate y-3 / y?
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\[\int\limits \frac{y-3}{y}dy?\]
Yes.
Separate the denominator
\[\int\limits 1-\frac{3}{y}dy\]
\[\int\limits 1-3y^{-1}dy\] Its just power rule now
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So I wil get y-3y..right?
Nope, integrating \(y^{-1}\) gives off ln(y)
It is y - 3 In y?
Careful, you forgot the constant
Is it correct?
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Nope, I told you already you forgot the constant. You will get a constant whenever you do indefinite integrals
So 1-3 In y?The constant is 1,right?Is it?
No the constant is "c" \[\int\limits\limits 1-3y^{-1}dy\] \[=y-3\ln(y)+c\]
Okay,I got it. But if the question also have dx and dy,where must I put C?
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