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
