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)

Here specifically is what you did wrong. Because you changed the limits of integration from $(-3, 0)$ to $(21, 3)$, therefore once you found the antiderivative $-{1\over u}$, you should have just plugged in $-{1\over 21}$ and $-{1\over 3}$, not $-{1\over 2(21)^3+3}$ and $-{1\over 2(3)^3+3}$.
