\textbf{Quadratic trinomial}

How do you solve \[\int\frac{\mathrm{d}x}{ax^2 + bx + c}\] and \[\int\frac{\mathrm{d}x}{\sqrt{ax^2 + bx + c}}\] ? You write \[ax^2 + bx + c\] in the following way (hint: completing the square by adding and subtracting \[\frac{b^2}{4})\]:

,align \qquad ax^2 + bx + c &= a\left(x^2 + \frac{b}{a}x + \frac{c}{a}\right) = a\left(\left(x^2 + 2\frac{b}{2a}x + \frac{b^2}{4a^2}\right) +\left(- \frac{b^2}{4a^2}  + c \right)\right)\\&= a\left(\left(x + \frac{b}{2a}\right)^2 +\left(\sqrt{c - \frac{b}{2a}}\right)^2\right)\\&=a(t^2 + k^2),\quad t= x+ \frac{b}{2a},\quad k=\sqrt{c-\frac{b}{2a}}

Now the integral reduces to either  \[\boxed{\frac{1}{a}\int\frac{\mathrm{d}x}{t^2 + k^2} = \frac{1}{ak}\arctan{\frac{t}{k}} + C}\] or \[\boxed{\frac{1}{a}\int\frac{\mathrm{d}x}{\sqrt{t^2 + k^2}} = \frac{1}{a}\ln\left|t^2 + \sqrt{t^2 + k^2}\right| + C}\]