Advanced integration, how to integrate 1/polynomial ? Thanks I have been trying to integrate a function with a polynomial as the denominator.
i.e, how would I go about integrating 
$$\frac{1}{ax^2+bx+c}.$$
Any help at all with this would be much appreciated, thanks a lot :)
ps The polynomial has NO real roots 
$${}{}$$

If $a=0$, then
$$I = \int \dfrac{dx}{bx+c} = \dfrac1b \log (\vert bx+c \vert) + \text{constant}$$
$$I = \int \dfrac{dx}{ax^2 + bx + c} = \dfrac1a \int\dfrac{dx}{\left(x + \dfrac{b}{2a}\right)^2 + \left(\dfrac{c}a- \dfrac{b^2}{4a^2} \right)}$$
If $b^2 < 4ac$, then recall that
$$\int \dfrac{dt}{t^2 + a^2} = \dfrac1a\arctan \left(\dfrac{t}a \right) + \text{constant}$$
Hence, if $b^2 < 4ac$, then
$$I = \dfrac2{\sqrt{4ac-b^2}} \arctan \left( \dfrac{2ax + b}{\sqrt{4ac-b^2}} \right) + \text{constant}$$
If $b^2 = 4ac$, then
$$I =\dfrac1a \dfrac{-1}{\left(x + \dfrac{b}{2a}\right)} + \text{constant} = - \dfrac2{2ax+b} + \text{constant}$$
If $b^2 > 4ac$, then
$$I = \dfrac1a \int\dfrac{dx}{\left(x + \dfrac{b}{2a}\right)^2 - \sqrt{\left( \dfrac{b^2}{4a^2} -\dfrac{c}a\right)}^2}$$
Now $$\int \dfrac{dt}{t^2 - k^2} = \dfrac1{2k} \left(\int \dfrac{dt}{t-k} - \int \dfrac{dt}{t+k} \right) = \dfrac1{2k} \log \left(\left \vert \dfrac{t-k}{t+k} \right \vert \right) + \text{constant}$$
Hence, if $b^2 > 4ac$, then
$$I = \dfrac1{\sqrt{b^2-4ac}} \log \left(\left \vert \dfrac{2ax + b - \sqrt{b^2-4ac}}{2ax + b + \sqrt{b^2-4ac}} \right \vert \right) + \text{constant}$$
