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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- 2$\begingroup$ This is a megaduplicate. See for instance Getting different answers when integrating using different techniques $\endgroup$Anne Bauval– Anne Bauval2023-10-17 15:33:28 +00:00Commented Oct 17, 2023 at 15:33
- $\begingroup$ @SassatelliGiulio so it's all a matter of what constant I choose? $\endgroup$Dror– Dror2023-10-17 15:37:52 +00:00Commented Oct 17, 2023 at 15:37
- 2$\begingroup$ Yes, that's right. Your two approaches resulted in answers that only differed by a constant. $\endgroup$user2661923– user26619232023-10-17 15:40:50 +00:00Commented Oct 17, 2023 at 15:40
- 1$\begingroup$ Please 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. $\endgroup$CiaPan– CiaPan2023-10-17 16:20:14 +00:00Commented Oct 17, 2023 at 16:20
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1 Answer
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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 $$(1/2)(\ln|x+1|)+C_1.$$
