What is the probability of $X \le -\frac{1}{2a}$, given that the density function of $X$ is:
$$f_X(x)=\frac{1}{ax^2+x+1}$$
I had this question in a test. It turns out I should've circled $\frac{1}{2}$ but I actually integrated this, using partial fractions, and came with a function that for any $a$ in its domain, is NOT a CDF:
$$F_X(x)=\dfrac{1}{\sqrt{1-4a}}\ln{\left|\dfrac{2ax+1-\sqrt{1-4a}}{2ax+1+\sqrt{1-4a}}\right|}$$
$F_X$ is not a CDF because its limit at $\infty$ is not $1$ but $0$. Also, when I apply $x=-\frac{1}{2a}$ I get $\ln(1)=0$ too.
Can anyone shed some light on this?
Acutally, the integral is \begin{equation} \int \frac{1}{ax^2+x+1} dx = \frac{2\tan^{-1}(\frac{2ax+1}{\sqrt{4a-1}})}{\sqrt{4a-1}} + c \end{equation}
$c$ and $a$ can be get from assumptions of CDF in $-\infty, \infty$ (assuming that support is ($-\infty, \infty$)).
However more clever solution involves spotting that the distribution is symmetrical in relation to $-\frac{1}{2a}$.
This implies that the $F(-\frac{1}{2a}) = \frac{1}{2}$