1. ## Hard Integration Q

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

I've tried all sorts of substituors for this one but i still cant seem to work it out. Can someone help out on this one?

2. You could try saying $\displaystyle x^4$ = $\displaystyle tan^2 y$

$\displaystyle dx = (sec^2 y)/[4(tan y)^{3/2}]$

$\displaystyle x^2 = tan y$

Which would give you 1/4 times the integral of (tan y - 1)/[(tan y)^(3/2)(tan y + 1)]

I am not sure if that will lead any where and you may need another substitution.

3. not so sure about the final answer but if you set

$\displaystyle u = x^2$

you'll get

$\displaystyle \frac {1} {\sqrt{1 + u^2}}$

and i know the antiderivative of that is

$\displaystyle \arcsin u$

Good luck

4. $\displaystyle \frac{x^{2}-1}{\left( x^{2}+1 \right)\sqrt{1+x^{4}}}=\frac{x^{2}-1}{x\left( x^{2}+1 \right)\sqrt{x^{2}+\frac{1}{x^{2}}}}=\frac{1-\frac{1}{x^{2}}}{\left( x+\frac{1}{x} \right)\sqrt{\left( x+\frac{1}{x} \right)^{2}-2}},$

substitute $\displaystyle x+\frac1x=\frac{\sqrt2}t$ and your integral equals $\displaystyle -\frac1{\sqrt2}\arcsin\left(\frac{\sqrt2x}{1+x^2}\r ight)+k.$

5. Originally Posted by Krizalid
$\displaystyle \frac{x^{2}-1}{\left( x^{2}+1 \right)\sqrt{1+x^{4}}}=\frac{x^{2}-1}{x\left( x^{2}+1 \right)\sqrt{x^{2}+\frac{1}{x^{2}}}}=\frac{1-\frac{1}{x^{2}}}{\left( x+\frac{1}{x} \right)\sqrt{\left( x+\frac{1}{x} \right)^{2}-2}},$

substitute $\displaystyle x+\frac1x=\frac{\sqrt2}t$ and your integral equals $\displaystyle -\frac1{\sqrt2}\arcsin\left(\frac{\sqrt2x}{1+x^2}\r ight)+k.$
You show 'em son.

6. Wow, just wow. You are unbelievable!

Originally Posted by Krizalid
$\displaystyle \frac{x^{2}-1}{\left( x^{2}+1 \right)\sqrt{1+x^{4}}}=\frac{x^{2}-1}{x\left( x^{2}+1 \right)\sqrt{x^{2}+\frac{1}{x^{2}}}}=\frac{1-\frac{1}{x^{2}}}{\left( x+\frac{1}{x} \right)\sqrt{\left( x+\frac{1}{x} \right)^{2}-2}},$

substitute $\displaystyle x+\frac1x=\frac{\sqrt2}t$ and your integral equals $\displaystyle -\frac1{\sqrt2}\arcsin\left(\frac{\sqrt2x}{1+x^2}\r ight)+k.$

7. Originally Posted by vuze88
Find $\displaystyle \int \left ( \frac{x^2-1}{x^2+1}.\frac{1}{\sqrt{1+x^4}} \right )dx$

I've tried all sorts of substituors for this one but i still cant seem to work it out. Can someone help out on this one?

Or if you never think of the wonderful substitution $\displaystyle t = x - 1/x$ , you can choose this substitution $\displaystyle x = \tan(y)$ which is familiar with all of us .

$\displaystyle dx = \sec^2(y) dy$

the integral becomes :

$\displaystyle I = \int \frac{ \tan^2(y) - 1}{(\tan^2(y) + 1)\sqrt{1+ \tan^4(y)}} \sec^2(y)~ dy$

$\displaystyle = \int \frac{ \tan^2(y) - 1}{\sqrt{1+ \tan^4(y)}}~ dy$

$\displaystyle = \int \frac{ \cos^2(y) }{ \cos^2(y) }\cdot \frac{ \tan^2(y) - 1}{\sqrt{1+ \tan^4(y)}}~ dy$

$\displaystyle = \int \frac{ \sin^2(y) - \cos^2(y) }{ \sqrt{ \sin^4(y) + \cos^4(y) }} ~dy$

$\displaystyle = - \int \frac{ \cos(2y) }{ \sqrt{ (\sin^2(y) + \cos^2(y) )^2 - 2\sin^2(y) \cos^2(y) }}~dy$

$\displaystyle = - \sqrt{2} \int \frac{ \cos(2y) }{ \sqrt{ 2 - \sin^2(2y) }}~dy$

Haha , then substitute $\displaystyle \sin(2y) = t ~\implies 2\cos(2y)dy = dt$

$\displaystyle I = - \frac{1}{\sqrt{2}} \int \frac{dt}{\sqrt{2 - t^2}}$

$\displaystyle = - \frac{1}{\sqrt{2}}\arcsin(\frac{t}{\sqrt{2}}) + C$

$\displaystyle t = \sin(2y) = \frac{2\tan(y)}{ \tan^2(y) + 1 } = \frac{2x}{x^2+1}$

so

$\displaystyle I = - \frac{1}{\sqrt{2}}\arcsin(\frac{\sqrt{2}x}{x^2+1}) + C$

8. cool, thanks for that simplependulum

9. Originally Posted by simplependulum
Or if you never think of the wonderful substitution $\displaystyle t = x - 1/x$ , you can choose this substitution $\displaystyle x = \tan(y)$ which is familiar with all of us .

$\displaystyle dx = \sec^2(y) dy$

the integral becomes :

$\displaystyle I = \int \frac{ \tan^2(y) - 1}{(\tan^2(y) + 1)\sqrt{1+ \tan^4(y)}} \sec^2(y)~ dy$

$\displaystyle = \int \frac{ \tan^2(y) - 1}{\sqrt{1+ \tan^4(y)}}~ dy$

$\displaystyle = \int \frac{ \cos^2(y) }{ \cos^2(y) }\cdot \frac{ \tan^2(y) - 1}{\sqrt{1+ \tan^4(y)}}~ dy$

$\displaystyle = \int \frac{ \sin^2(y) - \cos^2(y) }{ \sqrt{ \sin^4(y) + \cos^4(y) }} ~dy$

$\displaystyle = - \int \frac{ \cos(2y) }{ \sqrt{ (\sin^2(y) + \cos^2(y) )^2 - 2\sin^2(y) \cos^2(y) }}~dy$

$\displaystyle = - \sqrt{2} \int \frac{ \cos(2y) }{ \sqrt{ 2 - \sin^2(2y) }}~dy$

Haha , then substitute $\displaystyle \sin(2y) = t ~\implies 2\cos(2y)dy = dt$

$\displaystyle I = - \frac{1}{\sqrt{2}} \int \frac{dt}{\sqrt{2 - t^2}}$

$\displaystyle = - \frac{1}{\sqrt{2}}\arcsin(\frac{t}{\sqrt{2}}) + C$

$\displaystyle t = \sin(2y) = \frac{2\tan(y)}{ \tan^2(y) + 1 } = \frac{2x}{x^2+1}$

so

$\displaystyle I = - \frac{1}{\sqrt{2}}\arcsin(\frac{\sqrt{2}x}{x^2+1}) + C$
You're a trooper for writing that out.