<s>[INST] Evaluate the integral $\int \frac{x}{(x^2 + 4)^5} \mathrm{d}x$ 
Evaluate the integral $$\int \frac{x}{(x^2 + 4)^5} \mathrm{d}x.$$

If I transfer $(x^2 + 4)^5$ to the numerator, how do I integrate? 
 [/INST] Put $u = x^2+4 \implies du = 2x\,dx\implies x\,dx = \frac 12 du$.
$$ \begin{align}\int \frac{x}{(x^2 + 4)^5} \mathrm{d}x &= \frac 12\int u^{-5} \,du\\ & = \frac 12 \left(\frac {u^{-4}}{-4}\right) +c \\ &= -\frac{1}{8u^4} + c \\ &= -\frac {1}{8(x^2 + 4)^4} + c\end{align} $$
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