Definite integral by u-substitution (1/u^2), u given $$\int_{-3}^0 \frac{-8x}{(2x^2+3)^2}dx; u=2x^2+3$$
I need help solving this integral -- I'm completely bewildered. I've attempted it many times already and I don't know what I am doing wrong in my work, and I seem to be having the same problem with other definite integrals with the format $\frac{1}{u^2}$.
Here is how I've worked it out without success:
$$du=4xdx$$
$$-2\int_{21}^3\frac{1}{u^2}du=-2\int_{21}^3{u^{-2}}du$$
$$-2(\frac{-1}{2(3)^2+3}-(\frac{-1}{2(21)^2+3}))$$
$$-2(\frac{-1}{21}+\frac{1}{885})=\frac{192}{2065}$$
According to WA and the answer key to my homework I should have gotten $\frac{4}{7}$, a far cry from my answer. What am I doing wrong? (I am new to StackExchange by the way -- I apologize for poor formatting or tagging on my part)

Since the derivative of $2x^2 + 3$ is almost the numerator, we set $u = 2x^2 + 3$. Then $du = 4x dx$ or $dx = du/(4x)$. Then take the integral so,
$$-\int \frac{8x}{(2x^2 + 3)^2}dx = -\int \frac{2}{u^2} du = \frac{2}{u} + C = \frac{2}{2x^3 + 3} + C.$$
Therefore
$$-\int_{-3}^0 \frac{8x}{(2x^2 + 3)^2}dx = 2\left[\frac{1}{2x^2 + 3}\right]_{-3}^0 = 2\left(\frac{1}{3} - \frac{1}{21}\right) = 2\left(\frac{7}{21} - \frac{1}{21}\right)  = \frac{4}{7}.$$
