# Math Help - Hard integral.. or maybe just me being brainless

1. ## Hard integral.. or maybe just me being brainless

Hello folks,
i have this integral i wish i never met, because i'm not able to solve it in any method i know about.
Here's the "beast":
$\int \frac{t^2(t^2-1)}{(t^2+1)^2(3t^4-2t^2+3)}$

Hoping that any of you can give me any hint on how to go with it,
greetings (and excuse me for my poor english),
SkyWolf.

3. ## Re: Hard integral.. or maybe just me being brainless

Originally Posted by SkyWolf
Hello folks,
i have this integral i wish i never met, because i'm not able to solve it in any method i know about.
Here's the "beast":
$\int \frac{t^2(t^2-1)}{(t^2+1)^2(3t^4-2t^2+3)}$
Look at this webpage.
Be sure to click the "Show steps" tab.

4. ## Re: Hard integral.. or maybe just me being brainless

Thanks both, i already looked into wolfram way before posting here, but the solution there calls for $tanh^{-1}$ (hyperbolic tangent?) which is something we didn't touched at all in my math course (which is the at the one in first year so no previous ones). So, there must be something odd with all this.

If it can help understanding where the issue is, i'll post the initial form of the integral before substitutions:

$\int \frac{(cosx-(cosx)^3)}{(1+2(cosx)^2)}$

So.. How come i have to complete this integral without even knowning about $tanh^{-1}$? It is doable elseway?

SkyWolf

5. ## Re: Hard integral.. or maybe just me being brainless

Originally Posted by SkyWolf
Thanks both, i already looked into wolfram way before posting here, but the solution there calls for $tanh^{-1}$ (hyperbolic tangent?) which is something we didn't touched at all in my math course (which is the at the one in first year so no previous ones). So, there must be something odd with all this.

If it can help understanding where the issue is, i'll post the initial form of the integral before substitutions:

$\int \frac{(cosx-(cosx)^3)}{(1+2(cosx)^2)}$

So.. How come i have to complete this integral without even knowning about $tanh^{-1}$? It is doable elseway?

SkyWolf
My version doesn't show the inverse hyperbolic tangent function, instead it shows logarithms (which are an alternate equivalent form).

6. ## Re: Hard integral.. or maybe just me being brainless

Originally Posted by Prove It
My version doesn't show the inverse hyperbolic tangent function, instead it shows logarithms (which are an alternate equivalent form).
I too see no inverse hyperbolic tangent function at that website.

7. ## Re: Hard integral.. or maybe just me being brainless

My wolfram then is indeed crazy showing inverse hyperbolic tangents.
Anyways, so the way is to find complex radix and proceed like that? Will try and see what i get.

Greetings,
SkyWolf.