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Evaluate $$\int_{-1}^{2} \frac{-1}{x-3} + \frac{1}{x+2} + \frac{1}{6} dx$$

Since complex numbers haven't been treated, I can't use them. Using the sum rules:

$$\int_{-1}^{2} \frac{-1}{x-3} dx + \int_{-1}^{2}\frac{1}{x+2} dx + \int_{-1}^{2}\frac{1}{6} dx$$ = $$\left[- \ln(x-3) \right]_{-1}^2 +\left[ \ln(x+2) \right]_{-1}^2 + \frac{x}{6} $$

I could write it out, but well, the first term will already give me a negative logarithm...

Then I tried fumbling around a bit with the rules for logarithms, for example:

$$-\ln(x-3) + \ln(x+ 2) = -1\ln(x-3) + \ln(x+ 2) = \ln((x-3)^{-1}) + \ln(x+ 2) = \ln((x-3)^{-1}(x+ 2)) $$... well this is heading nowhere...

I'm stuck! What approach should I take to solve this?

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Your problem is that you have forgotten that $$\int \frac{1}{x}dx=\ln\color{red}{|}x\color{red}{|} \ne \ln x$$ And so you will not end up with the logarithm of a negative number.

If you remember this, then you will instead end up with $$[-\ln|x-3|]_{-1}^2+[\ln|x+2|]_{-1}^2+\bigg[\frac{x}{6}\bigg]_{-1}^2$$ Which can be easily evaluated.

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  • $\begingroup$ You've invented a new integral. Where in the world did you find that $\int(1/x)~dx=\ln|x|$? What you need here is to say that $$\int_{-1}^2 \frac{1}{x-3}~dx=-\int_{-1}^2 \frac{1}{3-x}~dx$$ $\endgroup$ – Cye Waldman Jul 29 '17 at 22:31
  • $\begingroup$ @Apeiron ...what two cases? Yes, this integral seems common enough. $\endgroup$ – Franklin Pezzuti Dyer Jul 31 '17 at 13:03

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