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Interesting Integral Illation (Posted on 2023-04-06) Difficulty: 3 of 5
Evaluate this definite integral:

 n
 ∫ x2*(1+x4)-1 dx
1/n

where n is a real number greater than 1.

See The Solution Submitted by K Sengupta    
Rating: 5.0000 (1 votes)

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Solution re: Some analytic progress - Final formula Comment 3 of 3 |
(In reply to Some analytic progress by Brian Smith)

So the arctan addition formula is arctan(A) + arctan(B) = arctan((A+B)/(1-A*B)) + k*pi, where k is an integer.  For small A,B then we have k=0.  Just playing around I inferred for this problem that I need k=1.

I don't have a good explanation other than adding +pi is what works for this problem.

So then: 1/(2*sqrt(2)) * (arctan[sqrt(2)*x+1] + arctan[sqrt(2)*x-1]) |{1/n, n}
will simplify to 1/(2*sqrt(2)) * (arctan[sqrt(2)*x/(1-x^2)]  |{1/n, n} + pi)

Then evaluating at the bounds gives us
1/(2*sqrt(2)) * (arctan[sqrt(2)*n/(1-n^2)] - arctan[sqrt(2)*(1/n)/(1-1/n^2)] + pi)

The second arctan just simplifies to the negative of the first, so then after simplifying we get a decently compact final answer of 
(pi - 2*arctan[(sqrt(2)*n)/(n^2-1)])/(2*sqrt(2))

Just a quick check, at n=5 this formula evaluates as (pi-2*arctan(sqrt(2)*5/24))/(2*sqrt(2)) = 0.90812. 
And then putting the original integral into Wolfram Alpha we get (pi-arctan(sqrt(2)*120/263))/(2*sqrt(2)) = 0.90812.  
This looks like Wolfram is simplifying 2*arctan(stuff) by using the arctan addition formula one more time.  And doing that gives an alternate form of 
(pi - arctan[(sqrt(2)*(n^3-n))/(n^4-4n^2+1)])/(2*sqrt(2))

  Posted by Brian Smith on 2023-04-07 18:30:43
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