MX2 Integration Marathon (1 Viewer)

Qeru · Jan 29, 2021

Since this thread is dead I thought I would post a question: Find: $\int \frac{1}{\sqrt{x\sqrt{x}-x^2}} dx$

Randomstudent123 · Jan 29, 2021

Is this correct?

Qeru · Jan 29, 2021

Randomstudent123 said:
View attachment 30150

Is this correct?

Yes well done guess it was too easy Try:

$\int \frac{1}{\sqrt[4]{x^4+1}} dx$

vernburn · Jan 29, 2021

Qeru said:
Since this thread is dead I thought I would post a question: Find: $\int \frac{1}{\sqrt{x\sqrt{x}-x^2}} dx$

I did this a slightly different way:
$I=\int \frac{1}{\sqrt{x\sqrt{x}-x^2}} dx$
$=\int \frac{1}{\sqrt{x}\sqrt{\sqrt{x}-x}} dx$
$=2\int \frac{1}{2\sqrt{x}\sqrt{\frac14-\left(\sqrt{x}-\frac12\right)^2}} dx$
You will notice this is in the form $\int \frac{f'(x)}{\sqrt{a^2-\left[f(x)\right]^2}}dx=\sin^{-1}\left(\frac{f(x)}{a}\right)+C$ , yielding:
$2\sin^{-1}\left(2\sqrt{x}-1\right)+C$
I thought I had gotten it wrong at first because it differed from @Randomstudent123 's answer but I checked on Desmos and the graphs differ by $\pi$ and so it is indeed correct.

vernburn · Jan 30, 2021

Qeru said:
Yes well done guess it was too easy Try:

$\int \frac{1}{\sqrt[4]{x^4+1}} dx$

This is my first attempt but the answer‘s slightly off. On the line marked $(\ast)$ , if the denominator is $u^4-1$ instead of $1-u^4$ then I think the final answer is correct but I can’t find the missing negative.

notme123 · Jan 30, 2021

vernburn said:
This is my first attempt but the answer‘s slightly off. On the line marked $(\ast)$ , if the denominator is $u^4-1$ instead of $1-u^4$ then I think the final answer is correct but I can’t find the missing negative.

I dont think you converted from theta to x correctly. Shouldn't $(\sqrt{sin(theta)})=\frac{x}{\sqrt[4]{(x^4+1)}}$ , you did the reciprocal I think. Other than that you got it right.

notme123 · Jan 30, 2021

Qeru said:
Yes well done guess it was too easy Try:

$\int \frac{1}{\sqrt[4]{x^4+1}} dx$

Also @Qeru pls post more I was SOOOO close to getting $\int\frac{dx}{\sqrt[4]{x^4+1}$ . I used the right sub but my mind just died when I got to the part right before u-sub. I did the right u-sub as well I just did a dumb mistake with the differentials as its weird to work with $d(Theta)$ and $du$

vernburn · Jan 30, 2021

notme123 said:
I dont think you converted from theta to x correctly. Shouldn't $(\sqrt{sin(theta)})=\frac{x}{\sqrt[4]{(x^4+1)}}$ , you did the reciprocal I think. Other than that you got it right.

Yes, you are correct. I’m so dumb!

Qeru · Jan 30, 2021

Evaluate:
$\int_{0}^{\infty}\frac{\ln{x}}{1+x^2} dx$

vernburn · Jan 30, 2021

Qeru said:
Evaluate:
$\int_{0}^{\infty}\frac{\ln{x}}{1+x^2} dx$

Firstly, use the substitution $x=\frac{1}{u}\implies dx = -\frac{1}{u^2}du$ :
$I = -\int_{\infty}^{0}\frac{\ln{\frac{1}{u}}}{1+\frac{1}{u^2}} \cdot\frac{du}{u^2}$
$=\int_0^{\infty}\frac{-\ln{u}}{u^2+1}du$
$=-I$
Thus,
$2I=0\implies I=0$

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

MX2 Integration Marathon (1 Viewer)

Well-Known Member

New Member

Well-Known Member

Active Member

Active Member

Attachments

Well-Known Member

Well-Known Member

Active Member

Well-Known Member

Active Member

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

Users Who Are Viewing This Thread (Users: 0, Guests: 1)