I am getting 2 different answers for $\int \frac{1}{2x} \, dx $ and I do not know which is wrong. This is the first answer with steps: $$\int \frac{1}{2x} \, dx $$ $$\frac{1}{2}\int \frac{1}{x} \, dx $$ $$\frac{1}{2}\ln{|x|} +C $$ This is the second answer with steps: $$\int \frac{1}{2x} \, dx $$ $$\int1 (2x)^{-1} \, dx $$ $$u=2x $$ $$\frac{du}{dx}=2 $$ $$\frac{du}{2}=dx $$ $$\int1 (u)^{-1} \, \frac{du}{2} $$ $$\frac{1}{2}\int(u)^{-1} \, du $$ $$\frac{1}{2}\ln{|u|} +C $$ $$\frac{1}{2}\ln{|2x|} +C $$
Asked
Active
Viewed 101 times
1
-
1Possible duplicate of Is $\int \frac{1}{2x} , dx $ equal to $\frac{\ln|2x|}{2}+ C$ or $\frac{\ln|x|}{2}+ C$ – found instantly with Approach0 – Martin R Jul 31 '19 at 06:25
-
Also: https://math.stackexchange.com/q/820747/42969, https://math.stackexchange.com/q/1780251/42969, https://math.stackexchange.com/q/2037504/42969. – Martin R Jul 31 '19 at 06:29
-
$$(\log ax)'=\frac a{ax}=\frac1x$$ or $$(\log ax)'=(\log x+\log a)'=\frac1x.$$ – Jul 31 '19 at 06:51
2 Answers
0
They are the same. Just $\ln|2x|=\ln2+\ln|x|$ and $\frac{1}{2}\ln2$ goes in $C$.
Michael Rozenberg
- 194,933
0
Note that $$\ln|2x|=\ln 2|x|=\ln 2+\ln |x|,$$ so they differ only by a constant.
Hagen von Eitzen
- 374,180