[Integrales Indefinidas] Integrar, integrado una función con varios radicales

Integrar $\int\dfrac{\sqrt{3+x^2}-\sqrt{3-x^2}}{\sqrt{9-x^4}}dx$.

Solución:

Operamos el integrando
$$\dfrac{\sqrt{3+x^2}-\sqrt{3-x^2}}{\sqrt{9-x^4}}=\dfrac{\sqrt{3+x^2}-\sqrt{3-x^2}}{\sqrt{3+x^2}\sqrt{3-x^2}}=\dfrac{1}{\sqrt{3-x^2}}-\dfrac{1}{\sqrt{3+x^2}}$$
Reemplazamos la integral dada y operamos
$$\begin{align}\int\dfrac{\sqrt{3+x^2}-\sqrt{3-x^2}}{\sqrt{9-x^4}}dx&=\int\left(\dfrac{1}{\sqrt{3-x^2}}-\dfrac{1}{\sqrt{3+x^2}}\right)\\ &=\int\dfrac{1}{\sqrt{3-x^2}}dx-\int\dfrac{1}{\sqrt{3+x^2}}dx\\ &=\text{arc}\sin\left(\dfrac{x}{\sqrt{3}}-\ln\left|x+\sqrt{3+x^2}\right|\right)+c\end{align}$$