December 17th, 2017, 07:53 PM  #11 
Senior Member Joined: Sep 2015 From: USA Posts: 1,975 Thanks: 1026 
Yes, you can put me on your ignore list. Bai.

December 26th, 2017, 07:05 AM  #12 
Newbie Joined: Dec 2017 From: Netherlands Posts: 21 Thanks: 0  Leibniz's Rule
Here is a propose solution.

December 26th, 2017, 09:19 AM  #13 
Math Team Joined: Jan 2015 From: Alabama Posts: 3,191 Thanks: 871 
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 $F'(x)= \frac{4x^2+ 1}{x^4+ 1}\frac{4x^3(4x^2+ 1) 8x(x^4+ 1)}{(4x^2+ 1)^2}$. $F'(2)= \frac{17}{17}\frac{32(17) 16(17)}{17^2}= \frac{16}{17}$ Well done! (I strongly suggest you do NOT block romsek. He thought, from you first post, that you were asking us to do a "take home" test for you and, after realizing that you were practicing for a test, apologized. If you block romsek you will only prevent yourself from getting help from one of the best here.) Last edited by Country Boy; December 26th, 2017 at 09:25 AM. 
December 26th, 2017, 11:52 AM  #14 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,814 Thanks: 1046 Math Focus: Elementary mathematics and beyond 
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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