×

# 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
3 years, 3 months ago

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold
- bulleted- list
• bulleted
• list
1. numbered2. list
1. numbered
2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in $$...$$ or $...$ to ensure proper formatting.
2 \times 3 $$2 \times 3$$
2^{34} $$2^{34}$$
a_{i-1} $$a_{i-1}$$
\frac{2}{3} $$\frac{2}{3}$$
\sqrt{2} $$\sqrt{2}$$
\sum_{i=1}^3 $$\sum_{i=1}^3$$
\sin \theta $$\sin \theta$$
\boxed{123} $$\boxed{123}$$

Sort by:

@Anastasiya Romanova: I want to learn from you!

- 3 years, 3 months 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.

- 3 years, 3 months 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!

- 3 years, 3 months ago

@Anastasiya Romanova Madam can you suggest me any good book to increase my problem solving ability in calculas

- 3 years, 3 months ago