Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

$$\int\frac{x^{2}-1}{x^{3}\sqrt{2x^{4}-2x^{2}+1}} \: \text{d}x$$

I tried to substitute $x^2=t$ but I am unable to solve it and I also tried to divide numerator and denominator by $x^2$ and do something but could not get anything.

share|cite|improve this question
up vote 20 down vote accepted

Substitute $u=1/x$ to get $$ \int \frac{u^3 - u}{\sqrt{u^4-2u^2+2}}\,du $$ This integral is much simpler, and can be solved by substituting $v = u^4 - 2u^2 + 2$. The final result is $$ \int\frac{x^{2}-1}{x^{3}\sqrt{2x^{4}-2x^{2}+1}}\,dx \;=\; \frac{\sqrt{2x^4-2x^2+1}}{2x^2} + C. $$

share|cite|improve this answer
A well-deserved +1 :) – Chris Taylor Jun 15 '11 at 9:14

The factor of $(2x^4 - 2x^2 + 1)^{-1/2}$ suggests that it might be profitable to look at solutions of the form $f(x)(2x^4 - 2x^2 + 1)^{1/2}$, and hope for a simplification. Indeed, by differentiating this expression we get the differential equation

$$(2x^4 - 2x^2 + 1) \frac{\mathrm{d}f}{\mathrm{d}x} + 2x(2x^2 - 1) f = \frac{x^2-1}{x^3}$$

which can be solved with an integrating factor.

share|cite|improve this answer

Well, using mathematica I can see that this function the given answer is the derivative of the integral.


\begin{align*} f(x) &= \frac{\sqrt{2x^{4}-2x^{2}+1}}{x^{2}} \\ &= \frac{\frac{x^{2} \cdot (8x^{3}-4x)}{2 \sqrt{2x^{4}-2x^{2}+1}} - 2x \cdot \sqrt{2x^{4}-2x^{2}+1}}{x^{4}} \quad \ \Bigl[ \text{Note this is} \ f'(x)\Bigr] \\ &= \frac{x^{2} \cdot (4x^{3}-2x) - (2x^{4}-2x^{2}+1) \cdot 2x}{x^{4} \cdot \sqrt{2x^{4}-2x^{2}+1}} \\ &= \frac{4x^{4} -2x^{2} -4x^{4} + 4x^{2}-2}{x^{3} \cdot \sqrt{2x^{4}-2x^{2} +1}} \quad \Bigl[ \text{cancelling out x}\Bigr] \\ &= \frac{2 \cdot (x^{2}-1)}{x^{3} \cdot \sqrt{2x^{4}-2x^{2}+1}} \end{align*}

So write your integral as $$\int\frac{x^{2}-1}{x^{3}\cdot \sqrt{2x^{4}-2x^{2}+1}} = \frac{1}{2} \int \frac{\text{d}}{\text{dx}}\biggl(\frac{\sqrt{2x^{4}-2x^{2}+1}}{x^{2}}\biggr)\ \text{dx}$$

Using this you can get the answer.

share|cite|improve this answer

If all else fails, Wolfram Alpha can guide you through a solution step by step, although it is ridiculously complicated. I am sorry, but I do not see the solution immediately.

Click here for Wolfram Alpha evaluation - with solution.

share|cite|improve this answer
Oh dear thats really complicated. – Listing Jun 15 '11 at 8:42
Holy smoke, I didn't expect something that long...! – Hans Lundmark Jun 15 '11 at 17:10

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.