<s>[INST] Integral calculating issue If 
$$ \int \frac{1}{x+1}\,\mathrm{d}x = \ln|x+1|+C, $$ 
then why
$$ \int \frac{1}{x+\frac{1}{3}}\,\mathrm{d}x \not= \ln\left|x+\frac{1}{3}\right|+C? $$
 [/INST] $$\int\frac{f'}{f}=\ln(f)$$
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):
$$\ln(3x+1)-\ln(x+1/3)=\ln\left(\frac{3x+1}{x+1/3}\right)=\ln\left(\frac{3x+1}{x+1/3}\frac{3}{3}\right)=\ln\left(3\frac{3x+1}{3x+1}\right)=\ln(3)$$
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