Integrate $\int \frac{(1+x^2)dx}{(1−x^2)\sqrt{1+x^4}}$

Question

Integrate $\displaystyle \int \frac{(1+x^2)dx}{(1−x^2)\sqrt{1+x^4}}$

I don't know how to do this one. I need some suggestions. Thank you!

http://www.wolframalpha.com/input/?i=%5Cint+%5Cfrac%7B%281%2Bx%5E2%29dx%7D%7B%281%E2%88%92x%5E2%29%5Csqrt%7B1%2Bx%5E4%7D%7D; Should that be the answer to my integral? (I don't have the solutions, though) — L_McClain, Jan 09 '14 at 20:29
Are you sure you've copied the problem correctly? Your other calculus questions to date have been relatively routine, but this one is not (as your wolframalpha answer indicates). Life would be a lot easier with an extra $x$ in the numerator. — Barry Cipra, Jan 09 '14 at 20:48
@BarryCipra this problem is copied correctly, but what I don't know is what wolframalpha actually calculated ... — L_McClain, Jan 09 '14 at 20:51
Mathematica gives $$ \sqrt[4]{-1} \left(F\left(\left.i \sinh ^{-1}\left(\sqrt[4]{-1} x\right)\right|-1\right)-2 \Pi \left(i;\left.\sin ^{-1}\left((-1)^{3/4} x\right)\right|-1\right)\right) $$ where the special functions are the EllipticF and EllipticPi functions. — robjohn, Jan 09 '14 at 21:06
Mathematica does not handle these types of integrals very well, for some reason. This has been the case since at least version 6. — heropup, Jan 09 '14 at 21:08
@heropup: I have noticed that Mathematica has not handled several of the integrals that I've done for answers recently. However, with most of mine, it has just plain given up, rather than giving an answer that it cannot simplify. — robjohn, Jan 09 '14 at 21:13

heropup · Accepted Answer · 2014-01-09T21:20:05.710

8

These types of integrals are best evaluated using a substitution of the form $$u = x - x^{-1}, \quad du = 1 + x^{-2} \, dx.$$ Note that $u^2 = x^2 + x^{-2} - 2$, and we can write the integrand as $$\frac{1+x^2}{(1-x^2)\sqrt{x^4+1}} dx = \frac{x^2(1+x^{-2}) \, dx}{x^2(x^{-1}-x)\sqrt{x^2+x^{-2}}} = -\frac{du}{u\sqrt{u^2+2}} = -\frac{u \, du}{u^2 \sqrt{u^2+2}}.$$ Now let $v = \sqrt{u^2+2}$, $dv = \frac{u}{\sqrt{u^2+2}} \, du$, and the given integral becomes $$\begin{align*} \int \frac{dv}{2-v^2} &= \frac{\log(v+\sqrt{2}) - \log(v-\sqrt{2})}{2\sqrt{2}} + C \\ &= \frac{1}{2\sqrt{2}} \log \left| \frac{\sqrt{x^2+x^{-2}} + \sqrt{2}}{\sqrt{x^2+x^{-2}} - \sqrt{2}} \right| + C \\ &= \frac{1}{2\sqrt{2}} \log \left| \frac{\sqrt{x^4+1}+\sqrt{2}x}{\sqrt{x^4+1}-\sqrt{2}x} \right| + C. \end{align*}$$

edited Jan 09 '14 at 21:20

answered Jan 09 '14 at 21:01

heropup

135,869

2

There seems to be a factor of $-1$ missing, otherwise, this is correct. (+1) – robjohn Jan 09 '14 at 21:12
How do you know what you need to use as $u$? I wouldn't be able to see that myself and it's amazing what you did. – L_McClain Jan 09 '14 at 21:20
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@L_McClain: usually, it is experience. – robjohn Jan 09 '14 at 21:21
@robjohn: Thought so :) – L_McClain Jan 09 '14 at 21:22
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It's not a substitution that is commonly taught, as it tends to only apply to integrands of a very specific type. The only thing I can say is that once you have seen it, you will know. – heropup Jan 09 '14 at 21:22
That's true. :) – L_McClain Jan 09 '14 at 21:23
@heropup: I don't understand how you got this part: $\int \frac{dv}{2-v^2} = \frac{\log(v+\sqrt{2}) - \log(v-\sqrt{2})}{2\sqrt{2}} + C \ $ Is there a formula to do that or a rule? – L_McClain Jan 09 '14 at 21:41
$$\frac{1}{a^2-v^2} = \frac{1}{(a-v)(a+v)} = \frac{1}{2a}\left( \frac{1}{v+a} - \frac{1}{v-a}\right),$$ for $a > 0$. – heropup Jan 09 '14 at 21:53

score 3 · Answer 2 · answered Jan 09 '14 at 20:56

The integral you have is equivalent to, $$ \int \frac{1+x^2}{1-x^2} \frac{\,dx}{x\sqrt{x^2 + x^{-2} -2 + 2}} $$ Set $u = x - x^{-1}.$ Then the integral reduces to $$ -\int \frac{\,du}{u\sqrt{u^2 + 2}}. $$ This looks more promising: $$ -\int \frac{\,du}{u^2 \sqrt{2u^{-2} + 1}}. $$ Set $v = u^{-1}.$

Integrate $\int \frac{(1+x^2)dx}{(1−x^2)\sqrt{1+x^4}}$

2 Answers2