# [✓] Solve this indefinite integral?

Posted 7 months ago
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 I evaluate $$\int \frac{1}{y (y-1)} \, dy$$ which I expect to give $$\log (y-1)-\log (y)$$ and I get $$\log (1-y)-\log (y)$$Is this a bug?
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Posted 7 months ago
 Is this a bug? No this is a correct anti-derivative because of In[2]:= Together[D[Log[1 \[Minus] y] \[Minus] Log[y], y]] Out[2]= 1/((-1 + y) y) Read about another correct anti-derivative here Doing an integral: From Mathematica Version 1.0 to 11.0
Posted 7 months ago
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Posted 7 months ago
 Thank you very much. Yes, it never occurred to me that D[Log[1 - y], y] could be the same as D[Log[y - 1], y]Sorry for not stopping to think.I guess the reason Mathematica chooses the first anti-derivative is because of its preference for putting terms in $y^0$ first.If I had been dealing with y's less than 1, I would have been happy with my result...getting negative numbers inside the logarithm was a bit of a shock (not that I was actually evaluating them, but I could see that for the part of the parameter space I was dealing with it was a problem).
 It's still good to have A. P. Prudnikov, Yu. A. Brychkov, O. I. Marichev (who works since years with Wolfram): Integrals and Series (Moscow 1981, russian) on the book shelf, and there in § 1.2.5 no. 9, p. 31: Integrate[1/(x (x + a)), x] = Log[Abs[x/(x+a)]]/a it's the same thing as discussed in Doing an integral ... mentioned above. To make a long story short: use the Prudnikov et. al. solution as long as $x$ is real.Two of the integral table workers (Brychkow and Marichev) wrote an interesting blog article New Derivatives of the Bessel Functions Have Been Discovered with the Help of the Wolfram Language! Enjoy!