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December 17th, 2017, 08:53 PM   #11
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December 26th, 2017, 08:05 AM   #12
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Leibniz's Rule

Here is a propose solution.
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December 26th, 2017, 10:19 AM   #13
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It's easy to check, by actually doing the integral and then differentiating it.
$F(x)= \int_{2x}^{x^2}\frac{2t}{t^2+ 1}dt$

Let $u= t^2+ 1$. Then $du= 2tdt$, When $t= 2x$, $u= 4x^2+ 1$ and when $t= x^2$, $u= x^4+ 1$, so the integral becomes
$F(x)= \int_{4x^2+ 1}^{x^4+ 1} \frac{du}{u}= \left[ ln(u)\right]_{4x^2+ 1}^{x^4+ 1}= ln\left(\frac{x^4+ 1}{4x^2+ 1}\right)$

The derivative of that is $F'(x)= \frac{4x^2+ 1}{x^4+ 1}\frac{4x^3(4x^2+ 1)- 8x(x^4+ 1)}{(4x^2+ 1)^2}$.

$F'(2)= \frac{17}{17}\frac{32(17)- 16(17)}{17^2}= \frac{16}{17}$

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Last edited by Country Boy; December 26th, 2017 at 10:25 AM.
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December 26th, 2017, 12:52 PM   #14
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The given integral has an easy antiderivative (ignoring the constant of integration). Compute that and then some manipulation according to log identities gets you an answer. Not mentally demanding at all, if you know as much as you should, but perhaps somewhat tedious.
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