This is a nasty time-consuming integral.
The suggested substitution eventually gives (16u-16u3)/[(3+u2)(1+u2)]...and you need to look up how that tangent substitution gets you this result...I can't give you all the details here.
That integral can be integrated by partial fractions and the result is the sum of 2 terms involving the natural log of the two terms in the denominator.
At least that is what I got! Good luck!
Paul M.
tutor
Doug: I am with you down to the line where you have the binomial cubed in the denominator above the line and 2+fraction below the line...then I lose you. I get u(1-u^2)/[(3+u^2)(1+u^2)^2]. If that is correct, there is a partial fraction decomposition, but it is certainly ugly & I don't want to attempt it. Please comment.
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10/11/24
Doug C.
Ahh, I multiplied numerator and denominator of the "big" fraction by (1 + t^2)^3 turning a complex fraction to a simple (not so simple:)) fraction. I added a row to the graph to confirm that your solution gives the same result--it does! desmos.com/calculator/lenoqcbogv So whatever strategy you used to get to your denominator was a better strategy than mine.
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10/11/24
Doug C.
And I see how you did that, getting a common denominator for the two terms in the denominator, then multiplying numerator of big fraction by reciprocal of the new denominator. Yes, that partial fraction decomp is going to be really ugly.
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10/11/24
Doug C.
And here is the solution with an exact value. desmos.com/calculator/ivil8669h0
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10/11/24
Paul M.
tutor
Thanks, Doug. I was also able to get the solution in closed form after a real nuisance partial fraction decomposition (6 unknowns!!) and integration...and it matched your solution numerically. As I said in my initial answer the closed form is too "ugly" to post here.
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10/12/24
Doug C.
I got something different, but am not sure of my work. And I have no idea if there is a way to find an antiderivative for the integrand shown here--since the denominator does not factor by partial fractions not an option? desmos.com/calculator/uzz2mu4qqm10/10/24