Calculate $\int\frac1{x^2+bx+c}dx$ where $b$ and $c$ are real numbers and $d=4c-b^2$ Let's assume $d>0$ as well.
$$\int\frac1{x^2+bx+c}dx$$
where $b$ and $c$ are real numbers and $d=4c-b^2$
So my observation here is that this looks mightily similar to the standard integral that results in $\arctan u$, that integral usually solves from some form like $\frac1{u^2+1}$ where u is whatever has been substituted. 
When the conditions are like this with the real numbers $b,c$. And $d$ being an expression. I'm not really sure where to even begin. Throws my integral intuition out the window!
Any ideas on how a solution is reached on this one? Much appreciated. :)

Complete the square, we have
$$x^2 + bx + c = \left(x - \frac b 2\right)^2-\left(\frac b 2\right)^2 + c$$
$$= \left(x - \frac b 2\right)^2-\frac {b^2} 4 + c$$
Applying $d = 4c - b^2$, 
$$\left(x-\frac b 2\right)^2 + \frac 1 4 \left(4c - b^2\right)=\left(x - \frac b 2\right)^2 +\frac d 4$$
So we have, 
$$\int \frac 1 {\left(x - \frac b 2\right)^2 + \frac d 4} \mathrm dx$$
Now we can apply a substitution of 
$$x - \frac b 2 = \frac{\sqrt d} 2\tan\theta$$
To evaluate the integral. You can probably take it from here. 
