If \(\int \frac{\sqrt{4+x^2}}{x^6} d x\) = \(\frac{A\left(4+x^2\right)^{3 / 2}\left(B x^2-6\right)}{x^5}+C\), then A is:

1
-120
2
\(-\frac{1}{120}\)
3
120
4
\(\frac{1}{120}\)

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