Calculate $\int\frac1{x^2+bx+c}dx$ where $b$ and $c$ are real numbers and $d=4c-b^2$

Question

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. :)

Answer 1

The purpose of $d$ is merely to simplify notation: $$x^2+bx+c=(x-b/2)^2+c-b^2/4=(x-b/2)^2+d/4$$ Thus the substitution $u=x-b/2$ removes the linear term in the denominator: $$\int\frac1{x^2+bx+c}\,dx=\int\frac1{u^2+d/4}\,du$$ The integral then falls easily: $$=\frac2{\sqrt d}\tan^{-1}\frac2{\sqrt d}u+K=\frac2{\sqrt d}\tan^{-1}\frac2{\sqrt d}\left(x-\frac b2\right)+K$$

Answer 2

write $$x^2+bx+c=\left(x+\frac{b}{2}\right)^2+c-\frac{b^2}{4}$$

Answer 3

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.