<s>[INST] Integrate $\int \frac{x^2-2}{(x^2+2)^3}dx$ $$\int \frac{x^2-2}{(x^2+2)^3}dx$$
What I did :
Method $(1) $ Re writing $x^2-2 = (x^2+2)-4 $ and partial fractions.
Method $(2) $ Substituting $x^2 = 2\tan^2 \theta $ 
Is there any other easy methods ? 
Some substitution  ? 
 [/INST] Another approach:
Let:
$$I(c)=\int \frac{x^2-2}{x^2+c} dx$$
$$=\int \frac{x^2+c-c-2}{x^2+c} dx$$
The above step is can be avoided by long division, if it you see it as coming out of the blue. Anyways continuing:
$$=\int \left(1-\frac{c+2}{x^2+c} \right) dx$$
$$=x-\frac{(c+2) \arctan (\frac{x}{\sqrt{c}})}{\sqrt{c}}+C_1$$
Whet $c$ and $C_1$ are two different things. $C$ here represents the constant gained when integrating. Now find
$$\frac{I''(c)}{2}+C_2=\frac{1}{2} \frac{\partial^2}{\partial c^2}  \left(x-\frac{(c+2) \arctan (\frac{x}{\sqrt{c}})}{\sqrt{c}}+C_1 \right)+C_2 =\int  \frac{x^2-2}{(x^2+c)^3} dx$$
Where $C_2$ is another constant gained when integrating. You can more specifically find:
$$\frac{I''(2)}{2}+C_2$$
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