Integral of $\frac{1}{2x}$.

Question

Using $u$-substitution, $$ \int \frac{1}{ax} \,\mathrm{d}x = \frac{\ln|ax|}{a}. $$ However, should you have instead factored out $\frac{1}{a}$ and integrated $\frac{1}{x}$, your result would be $$\frac{1}{a} \int \frac{1}{x} \,\mathrm{d}x = \frac{ \ln|x| }{a}. $$

So which is correct? My personal experience suggests that the latter method is superior, but without $u$-substitution, how would one go about integrating $\frac{1}{ax+b}$?

Answer 1

Using $u$-substitution: $u=ax$, so $du=a\ dx$ and $$\int\frac{1}{ax}dx=\frac{1}{a}\int\frac{1}{u}du=\frac{\ln|u|}{a}+C=\frac{\ln|ax|}{a}+C=\frac{\ln|x|}{a}+\frac{\ln|a|}{a}+C=\frac{\ln|x|}{a}+D$$ Putting the factor out: $$\int\frac{1}{ax}dx=\frac{1}{a}\int\frac{1}{x}=\frac{\ln|x|}{a}+C$$

To integrate $\frac{1}{ax+b}$, use $u=ax+b$, so again $du=a\ dx$ and $$\int\frac{1}{ax+b}dx=\frac{1}{a}\int\frac{1}{u}du=\frac{\ln|u|}{a}+C=\frac{\ln|ax+b|}{a}+C$$ (which may also be rewritten as $\frac{\ln|x+\frac{b}{a}|}{a}+D$)

Answer 2

They're both correct: $$\frac{\ln|ax|}{a}=\frac{\ln|a|+\ln|x|}{a} =\frac{\ln|x|}a+c.$$

Answer 3

Note that

$$\frac{\ln|ax|}{a}-\frac{ \ln|x| }{a}=\frac1a\ln\frac{|ax|}{|x|}=\frac1a\ln|a|=c$$