Integral calculating issue

Question

If

$$ \int \frac{1}{x+1}\,\mathrm{d}x = \ln|x+1|+C, $$

then why

$$ \int \frac{1}{x+\frac{1}{3}}\,\mathrm{d}x \not= \ln\left|x+\frac{1}{3}\right|+C? $$

Answer 1

$$\int\frac{f'}{f}=\ln(f)$$ So both of your answers are correct. But you can do the second this way too: $$\frac{1}{x+1/3}=\frac{1}{x+1/3}\frac{3}{3}=\frac{3}{3x+1}$$ And it's integral can be written as $$\ln(3x+1)$$ But the $2$ results are the 'same' (Their difference is only a constant): $$\ln(3x+1)-\ln(x+1/3)=\ln\left(\frac{3x+1}{x+1/3}\right)=\ln\left(\frac{3x+1}{x+1/3}\frac{3}{3}\right)=\ln\left(3\frac{3x+1}{3x+1}\right)=\ln(3)$$

Answer 2

the result is $$\ln\left|x+\frac{1}{3}\right|+C$$ since differentiating this result for $x$ we get $$\frac{1}{x+\frac{1}{3}}$$

Answer 3

In general: $$\int \frac{1}{x+a} dx = \ln{|x+a|}+C.$$