Solving integral $\int \frac{3x-1}{\left(x^2+16\right)^3}$ 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:


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

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

Hint: Alternately, evaluate $I(a)~=~\displaystyle\int\frac{3x-1}{x^2+a}~dx$ assuming $a>0$, and then try to express your integral in terms of $I''(16)$.
