May 11th, 2017, 02:51 PM  #1 
Newbie Joined: May 2017 From: Houston Posts: 6 Thanks: 0  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.. 
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. WolframAlpha: 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: WolframAlpha: 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. 
June 4th, 2017, 09:49 PM  #9 
Newbie Joined: May 2017 From: Houston Posts: 6 Thanks: 0  

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