I need to solve this one integral. 
$$\int \frac{3x-1}{\left(x^2+16\right)^3}$$

You need to use the method of undetermined coefficients. That's what I get:

![enter image description here](https://i.stack.imgur.com/mIOft.png)
![enter image description here](https://i.stack.imgur.com/p61db.png)

$$(3x-1) = (Ax + B)(x^{2}+16)^{2} + (Cx + D)(x^{2}+16) + (Ex + F)$$
$$1: 256B + 16D + F = -1$$
$$x: 256A + 16C + E = 3$$
$$x2: 32B + D = 0$$
$$x3: 32A + C = 0$$
$$x4: B = 0$$
$$x5: A = 0$$

$$A = 0;B = 0;C = 0;D = 0;E = 3;F = -1;$$
![enter image description here](https://i.stack.imgur.com/xSL4l.png)

It turns out that I'm back to the same integral. What is wrong I do?