# 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?

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Hint: make the substitution $u = 4x^{2} +1$. Then $du = 8xdx$, and the integral becomes:

$$\frac{1}{4} \int \frac{8xdx}{4x^{2}+1} = \frac{1}{4} \int \frac{du}{u}$$

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Try differentiating $\ln(4x^2+1)$ and see what you get (don't forget the chain rule!). Then see how this compares to the integrand.

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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

$$\frac{2x}{4x^2+1}=\frac14\frac{(4x^2+1)'}{4x^2+1}\ldots$$

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