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$$\int_{-3}^{-2} \frac{dx}{2x+1}$$

On solving this I'm getting $\log(-3) - \log(-5)$ and isn't negative log not possible?

M.Mass
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Dhruv Raghunath
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2 Answers2

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You can, as multiple people have suggested, simply use $\int \frac1x\mathrm{d}x = \ln\left|x\right|+C$ to get $\frac12\left(\ln3 - \ln5\right) = -\frac12\ln\frac53$, which is the correct value. However, it is worth noting that you don't need the absolute value in any way whatsoever to achieve this answer. Frankly, it is a simplification that makes things nice when dealing with the reals, but you can just as easily use $\int \frac1x\mathrm{d}x = \ln x+C$ and use the identity $\ln\left(-x\right) = \ln x + i\pi$ to get the same result. Of course, the $i\pi$'s will cancel since the arguments of $\ln$ at both bounds of integration are negative. At least in my opinion, this gives a better intuition for why and how this integral behaves this way.

EDIT: Since complex numbers might be a bit much for what is clearly an introductory calculus problem, note that we also do not need the aforementioned identity, assuming you know log rules. If you allow yourself the freedom to write and manipulate the expression $\frac12\left(\ln\left(-3\right)-\ln\left(-5\right)\right)$ with the negatives included, you can simply use log rules to get $-\frac12\ln\frac53$ as above.

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    While I like this, the complex logarithm (the identity you presented) isn't shown in introductory calculus classes. – Sean Roberson Jun 27 '17 at 19:39
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    I acknowledge this fully. Typically, I would have presented a more complete elementary approach to ensure that I'm actually being helpful, but it seems others beat me to the punchline on that one. It can never be a bad thing to get exposure to new ideas, though! –  Jun 27 '17 at 19:45
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HINT: $$\int\frac{1}{2x+1}dx=\frac{1}{2}\log|2x+1|+C$$

Saketh Malyala
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Dr. Sonnhard Graubner
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