MX2 Integration Marathon (2 Viewers)

vernburn · Jan 30, 2021

Here's one:
$\int \frac{x^2-1}{\left(x^2+1\right)\sqrt{x^4+1}}dx$

Qeru · Jan 30, 2021

vernburn said:
Here's one:
$\int \frac{x^2-1}{\left(x^2+1\right)\sqrt{x^4+1}}dx$

We can use the divide by $x^2$ on the top and bottom trick:

$\int \frac{x^2-1}{\left(x^2+1\right)\sqrt{x^4+1}}dx$

$I=\int \frac{1-\frac{1}{x^2}}{\left(x+\frac{1}{x}\right)\sqrt{\frac{x^4+1}{x^2}}}dx$

$I=\int \frac{1-\frac{1}{x^2}}{\left(x+\frac{1}{x}\right)\sqrt{(x+\frac{1}{x})^2-2}}dx$

Let $u=x+\frac{1}{x} \implies du=1-\frac{1}{x^2}dx$

$\therefore I=\int \frac{1}{u \sqrt{u^2-2}}du$

Let $s=\sqrt{u^2-2} \implies ds=\frac{u}{\sqrt{u^2-2}} du \implies du=\frac{\sqrt{u^2-2}}{u} ds$

$\therefore I=\int \frac{1}{u^2}ds=\int \frac{1}{s^2+2} ds$

$\therefore I=\frac{1}{\sqrt{2}}\tan^{-1}\left(\frac{s}{\sqrt{2}}\right)+C$

$I=\frac{1}{\sqrt{2}}\tan^{-1}\left(\frac{\sqrt{x^2+\frac{1}{x^2}}}{\sqrt{2}}\right)+C$

Qeru · Jan 30, 2021

This is a cool one: $\int \left(\frac{x^2-3x+\frac{1}{3}}{x^3-x+1}\right)^2 dx$

notme123 · Jan 30, 2021

Qeru said:
This is a cool one: $\int \left(\frac{x^2-3x+\frac{1}{3}}{x^3-x+1}\right)^2 dx$

reverse quotient rule:
so basically the numerator should be in the form vu'-uv' or $(x^3-x+1)u'-u(3x^2-1)$ since $v=x^3-x+1, v'=3x^2-1$ , and denominator is already in the form $v^2$
expanding the numerator $(x^2-3x+\frac{1}{3})^2 = x^4-6x^3+\frac{29x^2}{3}-2x+\frac{1}{9}$
since highest power is 4, u is a quadratic
$vu'-uv'=(x^3-x+1)(2ax+b)-(ax^2+bx+c)(3x^2-1)$
Therfore: $x^4-6x^3+\frac{29x^2}{3}-2x+\frac{1}{9}=(x^3-x+1)(2ax+b)-(ax^2+bx+c)(3x^2-1)$
Expanding and equating coefficients nets $a=-1, b=3, c=-\frac{26}{9}$
Therefore: the integral $\int \left(\frac{x^2-3x+\frac{1}{3}}{x^3-x+1}\right)^2 dx = \frac{-x^2+3x-\frac{26}{9}}{x^3-x+1} +C$

notme123 · Jan 30, 2021

Here's one I found, might not be overly difficult. Given $\sum_{n=1}^{\infty}\frac{1}{n!}=e-1$ , evaluate $\int_{0}^{1}{x\sqrt{x\sqrt[3]{x\sqrt[4]{x\sqrt[5]{...}}}}dx$

vernburn · Jan 30, 2021

notme123 said:
Here's one I found, might not be overly difficult. Given $\sum_{n=1}^{\infty}\frac{1}{n!}=e-1$ , evaluate $\int_{0}^{1}{x\sqrt{x\sqrt[3]{x\sqrt[4]{x\sqrt[5]{...}}}}dx$

Ahahahaha...

Look at Post #153 of this thread, page 8. It was the first post I ever made on BoS! The notation I use is a bit complicated and could be expressed more clearly though.

notme123 · Jan 30, 2021

vernburn said:
Ahahahaha...

Look at Post #153 of this thread, page 8. It was the first post I ever made on BoS! The notation I use is a bit complicated and could be expressed more clearly though.

hahaha, amazing I did not know that. tbh I hate to admit it but I just looked up 'cool integrals' and found this.

vernburn · Jan 30, 2021

notme123 said:
hahaha, amazing I did not know that. tbh I hate to admit it but I just looked up 'cool integrals' and found this.

Tbh I’m surprised I remembered doing it. I did that over a year ago.

notme123 · Jan 30, 2021

vernburn said:
Tbh I’m surprised I remembered doing it. I did that over a year ago.

It's more impressive than my first post which was about me complaining about not wanting to do study notes for economics lol.

Qeru · Jan 30, 2021

vernburn · Jan 30, 2021

Qeru said:
$\int \frac{x^n}{ \sum_{k=0}^{n} \frac{x^k}{k!}}dx$

No idea if this is correct. Definitely a hard one.

Qeru · Jan 30, 2021

vernburn said:
No idea if this is correct. Definitely a hard one.

Yup nice now for a nasty one:

$\int{ \sqrt{\csc^2{x}+\cot^2{x}} dx$

Directrix · Jan 30, 2021

vernburn said:
No idea if this is correct. Definitely a hard one.

Neat handwriting for a future doctor!

idkkdi · Jan 30, 2021

Qeru said:
Yup nice now for a nasty one:

$\int{ \sqrt{\csc^2{x}+\cot^2{x}} dx$

Sqrt 1+t^2/2t + 1-t^2/2t = sqrt 1/t
Trivial qeru forgot about tan half angle

idkkdi · Jan 30, 2021

idkkdi said:
Sqrt 1+t^2/2t + 1-t^2/2t = sqrt 1/t
Trivial qeru forgot about tan half angle

Actually that’s still very ugly to integrate rip

Qeru · Jan 30, 2021

idkkdi said:
Sqrt 1+t^2/2t + 1-t^2/2t = sqrt 1/t
Trivial qeru forgot about tan half angle

? its cosec^2x and cot^2x not cosecx and cotx not to mention u need the dx

idkkdi · Jan 30, 2021

Qeru said:
? its cosec^2x and cot^2x not cosecx and cotx not to mention u need the dx

Tan half angle may or may not simplify what is meant to be integrated. Dw about my scribbling lmao. Did actually forget about the ^2 haha.

vernburn · Jan 30, 2021

Okay here’s my answer but it looks very VERY wrong. Don’t mock me!

idkkdi · Jan 30, 2021

Qeru said:
? its cosec^2x and cot^2x not cosecx and cotx not to mention u need the dx

Just a question. The standard way to do this integral seems to look like a page of working at least.
Any short ways?

idkkdi · Jan 30, 2021

vernburn said:
Okay here’s my answer but it looks very VERY wrong. Don’t mock me!

I see lns looks pretty legit lol.

Make a Difference – Donate to Bored of Studies!

MX2 Integration Marathon (2 Viewers)

Active Member

Well-Known Member

Well-Known Member

Well-Known Member

Well-Known Member

Active Member

Well-Known Member

Active Member

Well-Known Member

Well-Known Member

Active Member

Attachments

Well-Known Member

Forklift Certified

Well-Known Member

Well-Known Member

Well-Known Member

Well-Known Member

Active Member

Attachments

Well-Known Member

Well-Known Member

Users Who Are Viewing This Thread (Users: 0, Guests: 2)