Integral $\int_0^\infty \arctan\left(\frac{3x^2}{4x^4+5x^2+2}\right) \frac{dx}{x}$ I tried to solve this integral on Brilliant earlier using an identity for improper integrals, but it turned out to be wrong. I did some research and found that this identity only works when our function has a "bounded anti-derivative" on our interval. Here is the integral in question:
$$\int_0^\infty \frac1x \arctan\left(\frac{3x^2}{4x^4+5x^2+2}\right) \mathrm dx$$
Here is where I found this identity:
https://brilliant.org/wiki/integration-tricks/ 
Scroll down to where it says "Inversions" to find it.
The question I had was why you had to have a bounded anti-derivative in order to make the substitution? I understand that the answer I received from doing this doesn't make sense (as I got the integral diverges), but I don't understand why. Could anyone shed some light on this? 

$$\frac{3x^2}{1\color{blue}{+}4x^4+5x^2+1}=\frac{4x^2\color{blue}{-}x^2}{1\color{blue}{+}(4x^2+1)(x^2+1)}=\frac{(4x^2+1)\color{blue}{-}(x^2+1)}{1\color{blue}{+}(4x^2+1)(x^2+1)}$$
$$\arctan\left(\frac{(4x^2+1)\color{red}{-}(x^2+1)}{1\color{red}{+}(4x^2+1)(x^2+1)}\right)=\arctan(4x^2+1)\color{red}{-}\arctan(x^2+1)$$ Now you might notice why there was not needed any $\frac1x$ substitution. Also your integral becomes a Frullani-type integral, so:
$$\int_0^\infty \frac{\arctan(4x^2+1)-\arctan(x^2+1)}{x}dx=\left(\arctan (0) -\arctan (\infty ) \right)\ln \left(\frac12 \right) =\frac{\pi}{2}\ln 2$$
