How to evaluate this indefinite integral?

shakgoku
Messages
28
Reaction score
0

Homework Statement



\int \frac{\sqrt{x^4-1}}{x^3}dx

Homework Equations





The Attempt at a Solution



tried substituting x^4 = \sec^2 \theta to get rid of square root but it was of no use because, I got another complex integral \int \frac{\sin^2\theta}{\cos^{\frac{3}{4}} \theta}d\theta

Is there an easy way to solve this problem?
 
Last edited:
Physics news on Phys.org
why can't u take x^3 inside square root. Then it will be simplified to
sqrt(1-x^2)/x which is easier to integrate
 
test!
The quadratic formula is
http://texify.com/$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$
 
Last edited by a moderator:
n.karthick said:
why can't u take x^3 inside square root. Then it will be simplified to
sqrt(1-x^2)/x which is easier to integrate

I don't get it. if we take x^3 inside we get,

\int \sqrt{\frac{x^4-1}{x^6}}dx=\int \sqrt{\frac{1}{x^2}-\frac{1}{x^6}}dx
 
I just realized I made a blunder in calculation.
That substitution yields
\int \frac{\sin^2\theta}{cos \theta}d \theta=\int \frac{1-\cos^2\theta}{\cos \theta} d\theta

and finally it is equal to,

\int \sec \theta d\theta-\int \cos\theta d\theta = \ln |x^2+\frac{\sqrt{1-x^2}}{x^2}|-\sqrt{1-x^2}+C

Which is the answer!
:peace:
 
Thread 'Use greedy vertex coloring algorithm to prove the upper bound of χ'
Hi! I am struggling with the exercise I mentioned under "Homework statement". The exercise is about a specific "greedy vertex coloring algorithm". One definition (which matches what my book uses) can be found here: https://people.cs.uchicago.edu/~laci/HANDOUTS/greedycoloring.pdf Here is also a screenshot of the relevant parts of the linked PDF, i.e. the def. of the algorithm: Sadly I don't have much to show as far as a solution attempt goes, as I am stuck on how to proceed. I thought...
Back
Top