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

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.

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4 Answers

up vote 19 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 $$ \int\frac{x^{2}-1}{x^{3}\sqrt{2x^{4}-2x^{2}+1}}\,dx \;=\; \frac{\sqrt{2x^4-2x^2+1}}{2x^2} + C. $$

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A well-deserved +1 :) –  Chris Taylor Jun 15 '11 at 9:14
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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.

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

Consider

\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.

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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.

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