Question

integral x^(3n-1)/(x^(2n)+1)^2 ?

Answer 1

@Spacelimbus ?

Answer 2

I will have to take a look at this one in a few minutes, seems to be hyperbolic, or trigonometric, but I have to think about a valueable substitution. I have to go AFK, but I will look at this in a moment, if it's still unsolved.

Answer 3

I believe I am pretty close on this one, but it still turns out to be pretty tricky, I choose u=tan(alpha) as a substitution, because tan(alpha)^2+1= sec(alpha)^2

Answer 4

did you manage to solve this one already yourself @atsitpfp ?

Answer 5

no :(

Answer 6

let me relog real quick and then show you a few of the imperative steps, unfortunately my computer bugs on the windows partition, I require some LaTeX to answer this question.

Answer 7

Alright here I am. My first guess is that this integral is trigonometric, it could be hyperbolic, but I'd prefer to start with an trigonometric attempt.

So I choose:
\[\Large x^n=\tan\alpha \]
Because \[\Large \tan^2\alpha +1 = \sec\alpha \]
In order to get nice forms of x, solve for it:
\[\Large x= \sqrt[n]{\tan\alpha} \]

Answer 8

So differentiate that fir:
\[ \Large \frac{dx}{d \alpha}= \frac{1}{n}( \tan\alpha)^{\frac{1-n}{n}}=\frac{1}{n} \sqrt[n]{\tan\alpha} \cdot (\tan \alpha )^{-1} \]

Answer 9

how can you tell it's trigonometric? is this the quality you aquire if you have a lot of experience in this?

Answer 10

A bit of both, I am a member on OS for quite a while now, thus I have solved quite a lot of integrals. In addition to the ones I had to do during my own academic studies.

However, you're right, at first such integrals are quite confusing and there isn't anything you can do but "observe" and guess for a structure.

If you want a good website that deals with such integrals, then there's paul's website.

Answer 11

http://tutorial.math.lamar.edu/Classes/CalcII/TrigSubstitutions.aspx

Answer 12

thanks, the website looks quite comprehensible

Answer 13

It's one of the best in my opinion. If you want to see the next few steps, I can also demonstrate them, but I think by here it's only a playing and plugging into with the substitutions:
\[\Large x^{3n-1}=x^{3n}\cdot x^{-1}= \frac{\tan^3 \alpha}{\sqrt[n]{\tan \alpha}} \]