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Recently I've found an article about the Feynman's integration technique. A technique for solving hard integral, that the usual methods do not aply, or the usual methods would take realy long to solve, but the article, despite beeing very concise about the technique itself, jumps on a few steps that are told to be elementary, such as the partial fraction decomposition of a intemediate integral.

Well, I've tried to do the partial fraction but every try I do, I get something worse than presented, I dont know if I'm doing something wrong, but, if possible, I would like a detailed resolution of the integral down below. To help I'm leting the site link Richard Feynman’s integral trick

The integral mentioned above is: $$I(x) = \int_0^1{\frac{ln(x+1)}{x^2+1} dx}$$ The problem treats it as a function of $\alpha$: $$f(\alpha) = \int_0^1{\frac{ln(\alpha x+1)}{x^2+1} dx}$$ and states that $f(1) = I(x)$, and, $f(0) = 0$

As suggested my problem is in this steps. $$1^{st}: \frac{\partial f(\alpha)}{\partial \alpha} = \frac{1}{{\alpha}^2+1}\int_0^1{\left(\frac{\alpha + x}{x^2+1} + \frac{\alpha}{\alpha x+1} \right)d\alpha}$$ And the $2^{nd}$ is when it calls: $$I(1) = \int_0^1{\frac{ln(\alpha+1)}{{\alpha}^2+1}d\alpha}$$

Thakfully, liuzp

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marked as duplicate by Brevan Ellefsen, Lord Shark the Unknown, Nosrati, Isaac Browne, Claude Leibovici integration Jul 24 '18 at 8:13

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  • $\begingroup$ I suggest being more specific about what step exactly you've having difficulty with you. Even the abridged solution in the article is relatively long, and explaining all of the omitted details is a time-consuming and redundant undertaking, if you're only missing a few of them. $\endgroup$ – Travis Jul 23 '18 at 3:20
  • $\begingroup$ After the diferentiation, the article splits the integral in the partial fraction form, this is the first time that I could not understand. The second time is by the end when the article calls an similar integral of $I(1)$ and passes to the other side of equality. $\endgroup$ – liuzp Jul 23 '18 at 3:33
  • $\begingroup$ Can I revive some old post? I think that I could solve my problem with a simple reply on the post of the $1^{st}$ coment, but, I'm afraid of digging down an old dead post. $\endgroup$ – liuzp Jul 23 '18 at 3:56
  • $\begingroup$ In the first part of the question your neclarity is in how to split the fraction? For the second part note that $\int_e^f f(y)dy =\int_e^f f(k) dk$ $\endgroup$ – Zacky Jul 23 '18 at 7:52
  • $\begingroup$ Yes. I was trying by the method of assigning values to x such that all but one coefficiente becomes 0, however I keep getting strange thins for A, B and C. $\endgroup$ – liuzp Jul 24 '18 at 15:35