Integrate $\frac{x^2- 4}{x^4 + 24x^2 +16}$

Question

$\displaystyle\int\dfrac{x^2- 4}{x^4 + 24x^2 +16} dx$

Attempt:

I have tried splitting and solving it.

I have also tried completing the square but these methods aren't helping here.

Dividing numerator and denominator by $x^2$ didn't help too.

How do I go about solving it?

Answer 1

Use partial fraction decomposition:$$\frac{x-4}{x^2+24x+16}=\frac{1+\sqrt2}{2\left(x+12+8\sqrt2\right)}+\frac{1-\sqrt2}{2\left(x+12-8\sqrt2\right)}$$and therefore$$\frac{x^2-4}{x^4+24x^2+16}=\frac{1+\sqrt2}{2\left(x^2+12+8\sqrt2\right)}+\frac{1-\sqrt2}{2\left(x^2+12-8\sqrt2\right)}.$$

Answer 2

The denominator is a sum of squares $$ (4x)^2 + (x^2+4)^2 = x^4+24x^2+16 $$ Using that to suggest a trig substitution, we get $$ \int\dfrac{x^2- 4}{x^4 + 24x^2 +16} \;dx = -\frac{1}{4}\arctan\left(\frac{4x}{x^2+4}\right) + C $$

(Of course, the problem is rigged. If we had some other numerator, this method would probably not help.)

Answer 3

A REMARK

Because of $\dfrac{d}{dx}\left(\dfrac{x^2+4}{4x}\right)=\dfrac{x^2-4}{4x^2}$ we can deduce $$\displaystyle\int\dfrac{x^2- 4}{x^4 + 24x^2 +16} dx=\displaystyle 4\int \frac{d}{dx}\left(\arctan\dfrac{x^2+4}{4x}\right)\tag 1$$ But similarly we can get $$\displaystyle\int\dfrac{x^2- 4}{x^4 + 24x^2 +16} dx=-\frac14\displaystyle\int \frac{d}{dx}\left(\arctan\dfrac{4x}{x^2+4}\right)\tag 2$$ So, despite the apparent mistake one can give the answer $(1)$ instead of GEdgar's answer $(2)$.

In other words $$\displaystyle 4\int \frac{d}{dx}\left(\arctan\dfrac{x^2+4}{4x}\right)=-\frac14\displaystyle\int \frac{d}{dx}\left(\arctan\dfrac{4x}{x^2+4}\right)$$