We know that $∫{1\over x} dx$ $=$ $ln \lvert x\rvert$ $+$ $c$
If the $x$ within the natural logarithm is always positive, for example $x$ $=$ $3x^2$ $+$ $5$, should $∫{x\over 3x^2+5} dx$ be expressed as
$\int{x\over 3x^2+5} dx ={1\over6}\ln \lvert 3x^2+5\rvert+ c$
OR
$\int{x\over 3x^2+5} dx={1\over6}\ln (3x^2+5)$ $+$ $c$
Would it be wrong to use absolute value brackets instead of brackets if the value inside is always positive? Is it necessary to take into account the possibility of $x$ being a complex number / an imaginary number and thus making $\ln (3x^2+5)$ negative?
Which expression is more accurate?
Usually, the convention for a logarithm as the answer to an indefinite integral is $\ln |z|+C$, but if $|z|$ is positive, $\ln (z) + C$ is appropriate. In general for indefinite integrals, if it assumed that any variable will be positive, the absolute value bars are not required and can be dropped - but for logarithms, I would still use parentheses if the expression is part of the logarithm.
In your case, either $$\int \dfrac {x \ dx}{3x^2 +5} = \frac {1}{6} \ln |3x^2 + 5|+C$$ or $$\int \dfrac {x \ dx}{3x^2 +5} = \frac {1}{6} \ln (3x^2 + 5)+C$$ is perfectly fine as an answer as $3x^2 + 5$ is positive for all $x$.
It's when you come to definite integrals where the absolute value bars may be necessary. If $x \leq 0$, $\ln x$ is undefined in the reals; but if we get a value inside the absolute value bars that evaluates to a positive real number, the absolute value bars are not needed.