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# Learn to Fly

The following integral $\mathbb{A} = \displaystyle \int_{2}^{3} \dfrac{(x+3)^2+2^2}{(x+2)^2+3^2} \partial x$ has a closed-form value of $a + \log \left (\dfrac{b}{c} \right ) - \dfrac{\tan^{-1} \left ( \frac{d}{e} \right )}{f}$ Evaluate $$a+b+c+d+e+f$$, respecting the coprime fractions.

This problem was vaguely inspired by Anastasiya Romanova's integrals. Keep it up!

Note by Guilherme Dela Corte
2 years, 1 month ago

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@Anastasiya Romanova: I want to learn from you! · 2 years, 1 month ago

Learn to fly where? I don't have any intention to fly anywhere. Haha

Don't learn from me, I'm still learning too. $$\quad\ddot\smile$$

Anyway, we can use general expression

$\mathbb{A}=\int_q^p\frac{(x+p)^2+q^2}{(x+q)^2+p^2}\,dx$

Making substitution $$y=x-q$$, we get \begin{align} \mathbb{A}&=\int_0^{p-q}\frac{(y+p+q)^2+q^2}{y^2+p^2}\,dy\\ &=\int_0^{p-q}\left[\frac{y^2+2(p+q)y+(p+q)^2+q^2}{y^2+p^2}\right]\,dy\\ &=\int_0^{p-q}\left[\frac{y^2}{y^2+p^2}+\frac{2(p+q)y}{y^2+p^2}+\frac{(p+q)^2+q^2}{y^2+p^2}\right]\,dy\\ &=\int_0^{p-q}\left[1-\frac{p^2}{y^2+p^2}+\frac{2(p+q)y}{y^2+p^2}+\frac{(p+q)^2+q^2}{y^2+p^2}\right]\,dy\\ &=\int_0^{p-q}\left[1+\frac{2(p+q)y}{y^2+p^2}+\frac{(p+q)^2+q^2-p^2}{y^2+p^2}\right]\,dy\\ &=\int_0^{p-q}\left[1+\underbrace{\frac{2(p+q)y}{y^2+p^2}}_{\large\color{red}{\text{set}\,u=y^2+p^2}}+\underbrace{\frac{2(pq+q^2)}{y^2+p^2}}_{\large\color{blue}{\text{set}\,y=p\tan\theta}}\right]\,dy\\ \end{align} The rest is trivial. · 2 years, 1 month ago

“I always wonder why birds choose to stay in the same place when they can fly anywhere on the earth, then I ask myself the same question.” ― Harun Yahya

I insist on learning from a learner!

Thanks a lot for the generic case answer! · 2 years, 1 month ago