How do I integrate the following? $\int{\frac{(1+x^{2})\mathrm dx}{(1-x^{2})\sqrt{1+x^{4}}}}$

Question

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

This was a Calc 2 problem for extra credit (we have done hyperbolic trig functions too, if that helps) and I didn't get it (don't think anyone did) -- how would you go about it?

Answer 1

It might not be elliptic after all... (unless I have made some mistake)

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

Let $\displaystyle u = x -\frac{1}{x}$.

Then $\displaystyle du = (1 + \frac{1}{x^2})dx$.

Now $$\frac{1+x^2}{(1-x^2)\sqrt{1+x^4}} = -\frac{x^2(1 + 1/x^2)}{x(x-1/x)\sqrt{x^2(x^2 + 1/x^2)}} = -\frac{1 + 1/x^2}{(x-1/x)\sqrt{(x - 1/x)^2 + 2}}$$

Thus

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

$$= -\int \ \frac{\mathrm{du}}{u \sqrt{u^2 + 2}}$$

Answer 2

Somewhat inspired by Moron's wonderful answer, I decided to see if a trigonometric solution would do the job.

Making the substitution $x=\cot\left(\frac{\theta}{2}\right)$, $\mathrm dx=\frac{\mathrm d\theta}{\cos\;\theta-1}$, we have

$$\int \frac{\mathrm d\theta}{\cos\;\theta(1-\cos\;\theta)\sqrt{1+\cot^4\frac{\theta}{2}}}$$

$$=\int \frac{\mathrm d\theta}{\cos\;\theta(1-\cos\;\theta)\sqrt{1+\left(\frac{1+\cos\;\theta}{1-\cos\;\theta}\right)^2}}$$

$$=\frac1{\sqrt{2}}\int \frac{\mathrm d\theta}{\cos\;\theta\sqrt{1+\cos^2\theta}}$$

which integrates to

$$\frac1{\sqrt{2}}\tanh^{-1}\frac{\sin\;\theta}{\sqrt{1+\cos^2\theta}}$$

Undoing the substitution, we get

$$\frac1{\sqrt{2}}\tanh^{-1}\left(x\sqrt{\frac{2}{x^4+1}}\right)$$

and it is easy to verify that the derivative of this last expression gives the original integrand.

Answer 3

Moron's and J.M.'s solutions are nice. Hopefully this solution is simpler.

Without loss of generality we may assume that $1\gt x\gt 0$. Put $x:=\sqrt{y}$, $1\gt y\gt 0$. Then we obtain $$ \int{\frac{1+x^2}{(1-x^2)\sqrt{1+x^4}}}\mathrm dx=\int{\frac{1+y}{2(1-y)\sqrt{1+y^2}\sqrt{y}}}\mathrm dy. $$ Introduce the new variable $$ t:=\frac{1+y}{1-y},\qquad 1\lt t \lt \infty. $$ Then we have $$ y=\frac{-1+t}{1+t}, $$ $$ \mathrm dy=\frac{2}{(1+t)^2}\,\mathrm dt. $$ Substituting back we obtain $$ \int{\frac{1+y}{2(1-y)\sqrt{1+y^2}\sqrt{y}}}\mathrm dy=\int{\frac{t}{2\sqrt{1+\left(\dfrac{-1+t}{1+t} \right)^2}\sqrt{\dfrac{-1+t}{1+t}}}\frac{2}{(1+t)^2}\,\mathrm dt} $$ $$ =\frac{1}{\sqrt{2}}\int{\frac{t}{\sqrt{t^4-1}}}\mathrm dt $$ $$ =\frac{1}{2\sqrt{2}}\ln(t^2+\sqrt{t^4-1})+C. $$ Putting back everything we obtain $$ \frac{1}{2\sqrt{2}}\ln\left(\frac{(1+x^2)^2+2\sqrt{2}x\sqrt{1+x^4}}{(1-x^2)^2}\right)+C. $$

Answer 4

What a surprise! Surfing the net, I found an almost same question on "hard integral" $$ \displaystyle \int \frac{x^2 - 1}{(x^2 + 1) \sqrt{x^4 + 1}} \, dx $$ from June 19, 2008.

Answer 5

\begin{align}\int \frac{(1+x^2)dx}{(1-x^2)\sqrt{x^4+1}} =&\int \frac{\frac{(1+x^2)dx}{(1-x^2)|1-x^2|}}{\sqrt{\frac{x^4+1}{(1-x^2)^2}}} =\int \frac{d(\frac x{|1-x^2|})}{\sqrt{1+\frac{2x^2}{(1-x^2)^2}}} =\frac1{\sqrt2}\sinh^{-1}\frac{\sqrt2 x}{|1-x^2|} \end{align}

Answer 6

$$\begin{align*} & \int\frac{1+x^2}{(1-x^2)\sqrt{1+x^4}}\,dx \\ &= \int\frac{y}{\sqrt{1+\frac{(y-1)^2}{(y+1)^2}}}\cdot\frac{dy}{(y+1)\sqrt{y^2-1}} \tag1\\ &= \frac1{\sqrt2} \int\frac{y}{\sqrt{y^4-1}}\,dy \\ &= \frac1{2\sqrt2} \int\frac{dz}{\sqrt{z^2-1}} \tag2 \\ &= \frac1{2\sqrt2} \tanh^{-1}\left(\frac z{\sqrt{z^2-1}}\right) + C \\ &= \frac1{2\sqrt2} \tanh^{-1}\left(\frac{y^2}{\sqrt{y^4-1}}\right) + C \\ &= \frac1{2\sqrt2} \tanh^{-1}\left(\frac{\left(\frac{1+x^2}{1-x^2}\right)^2}{\sqrt{\left(\frac{1+x^2}{1-x^2}\right)^4-1}}\right) + C \\ &= \frac1{2\sqrt2} \tanh^{-1}\left(\frac{(1+x^2)^2}{\sqrt{8x^2(1+x^4)}}\right) + C \end{align*}$$

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• $(1)$ : substitute $y = \dfrac{1+x^2}{1-x^2} \color{#0000006F}{\iff x=\sqrt{\dfrac{y-1}{y+1}} \implies dx = \dfrac{dy}{(y+1)\sqrt{y^2-1}}}$
• $(2)$ : substitute $z=y^2 \color{#0000006F}{\iff y=\sqrt z \implies dy=\dfrac{dz}{2\sqrt z}}$