So the first is obvious: $$\int \frac{1}{2x+2} dx = \frac{1}{2}\ln(2x+2) + C$$ Now, what if I do something like this: $$\int \frac{1}{2x+2} dx = \int \frac{1}{2(x+1)} dx = (\frac{1}{2}\int \frac{1}{x+1} dx) =\frac{1}{2}\ln(x+1) + C$$ Why can't I "take" the half out of the integral? (Sorry if the formatting is weird)
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2This is a megaduplicate. See for instance Getting different answers when integrating using different techniques – Anne Bauval Oct 17 '23 at 15:33
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@SassatelliGiulio so it's all a matter of what constant I choose? – Dror Oct 17 '23 at 15:37
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2Yes, that's right. Your two approaches resulted in answers that only differed by a constant. – user2661923 Oct 17 '23 at 15:40
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1Please recall, that $\ln(k\cdot X)=\ln k + \ln X.$ In your case, a multiplier $2$ from within logarithm jumps outside as a $\ln 2$ addend and then gets assimilated by the $C$ constant. – CiaPan Oct 17 '23 at 16:20
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$$\int [1/(2x+2)]dx=(1/2)(\ln|2x+2|)+C.$$ Further approaching we will get, $$(1/2)(\ln|x+1|)+(\ln|2|)/2+C.$$ Now we can write $$(\ln|2|)/2+C =C_1.$$ So, the final result will be $$(1/2)(\ln|x+1|)+C_1.$$
CiaPan
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