<s>[INST] Integrate $\int \frac{x^2}{(x^2+4)^2}$ $$\int \frac{x^2\;dx}{(x^2+4)^2}$$I suppose you must have to use trigonometric substitution, or something, but do not even know where to begin to solve this integral, guys, please help me!!
 [/INST] Integrate by parts:
$$\int\frac{x^2}{(x^2+4)^2}dx=\int x\frac{x}{(x^2+4)^2}dx=\frac{1}{2} \int x(\frac{-1}{x^2+4})'dx=\frac{1}{2}(\frac{-x}{x^2+4} +\int \frac{1}{x^2+4}dx)=\frac{1}{2}(\frac{-x}{x^2+4} +\frac{1}{2}\arctan\frac{x}{2}) +C$$
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