$$\int_{-3}^0 \frac{-8x}{(2x^2+3)^2}dx; u=2x^2+3$$
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)
$\endgroup$ 42 Answers
$\begingroup$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 $$-\int_{-3}^0 \frac{8x}{(2x^2 + 3)^2}dx = 2\left[\frac{1}{2x^2 + 3}\right]_{-3}^0 = 2\left(\frac{1}{3} - \frac{1}{21}\right) = 2\left(\frac{7}{21} - \frac{1}{21}\right) = \frac{4}{7}.$$
$\endgroup$ 1 $\begingroup$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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