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$$\int_{-3}^0 \frac{-8x}{(2x^2+3)^2}dx; u=2x^2+3$$

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)

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  • $\begingroup$ You didn't actually integrate the $u^{-2}$, you just substituted the values $3$ and $21$ into that function. $\endgroup$ – David Mar 21 '14 at 3:05
  • $\begingroup$ BTW formatting is fine, no need to apologise! If you want bigger brackets you could use \Bigl(...\Bigr), or if you want TeX to choose the size automatically, \left(...\right) $\endgroup$ – David Mar 21 '14 at 3:08
  • $\begingroup$ This brings in another thing here noteworthy: When you do a u-sub, you may want to take the limits along. That principle does not always work (in that case you would have to backsub) but here it works great. So integrate $\frac{1}{u^2}$ and determine your new upper and lower limit from your substitution. $\endgroup$ – imranfat Mar 21 '14 at 3:11
  • $\begingroup$ Thank you for the helpful comments! I think I did integrate it correctly, because the integration of $u^{-2}$ is $\frac{-1}{u}$. I just didn't show it in my work. $\endgroup$ – homura Mar 21 '14 at 3:41
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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 $$-\int_{-3}^0 \frac{8x}{(2x^2 + 3)^2}dx = 2\left[\frac{1}{2x^2 + 3}\right]_{-3}^0 = 2\left(\frac{1}{3} - \frac{1}{21}\right) = 2\left(\frac{7}{21} - \frac{1}{21}\right) = \frac{4}{7}.$$

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  • $\begingroup$ Thank you for the step by step explanation! You made it really obvious where I went wrong. (Which is a good thing) $\endgroup$ – homura Mar 21 '14 at 3:36
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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 $-{1\over 2(3)^3+3}$.

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  • $\begingroup$ Thank you for pointing out exactly where I went wrong! $\endgroup$ – homura Mar 21 '14 at 3:38

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