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!

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TopNewest@Anastasiya Romanova: I want to learn from you! – Guilherme Dela Corte · 2 years, 3 months ago

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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. – Anastasiya Romanova · 2 years, 3 months ago

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“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 YahyaI insist on learning from a learner!

Thanks

a lotfor the generic case answer! – Guilherme Dela Corte · 2 years, 3 months agoLog in to reply

@Anastasiya Romanova Madam can you suggest me any good book to increase my problem solving ability in calculas – Sandeep Rathod · 2 years, 3 months ago

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