Need help with $\int \dfrac{2x}{4x^2+1}$ We want$$\int \dfrac{2x}{4x^2+1}$$
I only know that $\ln(4x^2 + 1)$ would have to be in the mix, but what am I supposed to do with the $2x$ in the numerator?

Again, as in your past question, there's a general case here: if $\,f\,$ is derivable then
$$\int\frac{f'(x)}{f(x)}dx=\log|f(x)|+K$$
Here, we have 
$$\frac{2x}{4x^2+1}=\frac14\frac{(4x^2+1)'}{4x^2+1}\ldots$$
