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May 11th, 2017, 02:51 PM   #1
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Integral

Could you please solve that integral step by step

integral from 0 to 1 : ln(x^2+1)/(x+1)

Also, I attached as a photo..
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 May 12th, 2017, 09:28 AM #2 Member   Joined: Feb 2015 From: Southwest Posts: 96 Thanks: 24 I ran your problem through wolfram just to see if its easily solvable. Wolfram|Alpha: Computational Knowledge Engine It doesn't look simple, you may want to make sure you have it written correctly.
 May 12th, 2017, 07:07 PM #3 Newbie   Joined: May 2017 From: Houston Posts: 6 Thanks: 0 Yes, it is not simple.. And it has boundaries.. It has special solution, and that is what I want - if you are able to solve. - It should has something like that . Actually this question asked by our professor in college as a hard question, but he said that it has a trick, and solvable by some of the series.. You have to do a lot of times IBT(Integration By Part), and cut at somewhere, and find something with Series.. Answer is -π^2/48 + (3 log^2(2))/4 according to WolframAlpha: Wolfram|Alpha: Computational Knowledge Engine) . After first IBT I got at the integral side : integral 2x[ln(x+1)/(x^2+1)]dx. From 0 to 1 integral of ln(x+1)/(x^2+1) equals (ln(2)π)/8 but, because of I'm not an advance about IBT for definite integrals, I could not walk anymore.. Thanks in advance for your helps Misty
 May 20th, 2017, 01:56 AM #4 Newbie   Joined: May 2017 From: Houston Posts: 6 Thanks: 0 Yes, it is hard!! There should be some tricks to solve it, and it has boundaries.. Thanks..
 May 25th, 2017, 10:18 AM #5 Newbie   Joined: May 2017 From: Houston Posts: 6 Thanks: 0 Any idea?
 May 25th, 2017, 12:44 PM #6 Newbie   Joined: Dec 2016 From: Austin Posts: 10 Thanks: 0 Special Integral Do you still need this? I attempted a solution on paper, but stopped just before plugging in my bounds. It may take me a bit to type the solution up. Just let me know. Thanks.
 May 25th, 2017, 12:50 PM #7 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,488 Thanks: 887 Math Focus: Elementary mathematics and beyond I am interested in seeing a solution.
 May 27th, 2017, 09:46 AM #8 Global Moderator   Joined: Dec 2006 Posts: 17,160 Thanks: 1284 Try something like the following: $\displaystyle \int_0^1\frac{\ln(x^2 + 1)}{x + 1}dx = \int_0^1\frac{\ln(x - i)}{x + 1}dx + \int_0^1\frac{\ln(x + i)}{x + 1}dx$ then finding those two integrals by writing the integrand as a series (possibly after making a substitution) and integrating that term by term. Thanks from greg1313
June 4th, 2017, 09:49 PM   #9
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Quote:
 Originally Posted by thegrade Do you still need this? I attempted a solution on paper, but stopped just before plugging in my bounds. It may take me a bit to type the solution up. Just let me know. Thanks.
Yes I still need that..

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