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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. :)

Parcly Taxel
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99 Fishing
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3 Answers3

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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: $$=\frac2{\sqrt d}\tan^{-1}\frac2{\sqrt d}u+K=\frac2{\sqrt d}\tan^{-1}\frac2{\sqrt d}\left(x-\frac b2\right)+K$$

Parcly Taxel
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  • Thank you for the beautiful answer. Questions like this sadden me but also motivate me to keep studying. Never would my current intuition have thought to solve the problem this way. I can understand it but this would never occur to me haha. – 99 Fishing Sep 18 '17 at 12:35
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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.

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

Dr. Sonnhard Graubner
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