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

How do I integrate the following:
$\int \frac{1+{x}^{2}}{\left(1-{x}^{2}\right)\sqrt{1+{x}^{4}}}dx$
You can still ask an expert for help

• Questions are typically answered in as fast as 30 minutes

Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it

Stuart Rountree
$\int \frac{1+{x}^{2}}{\left(1-{x}^{2}\right)\sqrt{1+{x}^{4}}}dx$
Let $u=x-\frac{1}{x}$
Then $du=\left(1+\frac{1}{{x}^{2}}\right)dx$
Now
$\frac{1+{x}^{2}}{\left(1-{x}^{2}\right)\sqrt{1+{x}^{4}}}=-\frac{{x}^{2}\left(1+\frac{1}{{x}^{2}}\right)}{x\left(x-\frac{1}{x}\right)\sqrt{{x}^{2}\left({x}^{2}+\frac{1}{{x}^{2}}\right)}}=-\frac{1+\frac{1}{{x}^{2}}}{\left(x-\frac{1}{x}\right)\sqrt{{\left(x-\frac{1}{x}\right)}^{2}+2}}$
Thus
$\int \frac{1+{x}^{2}}{\left(1-{x}^{2}\right)\sqrt{1+{x}^{4}}}dx$
$=-\int \frac{du}{u\sqrt{{u}^{2}+2}}$
###### Not exactly what you’re looking for?
Elaine Verrett
karton

Without loss of generality we may assume that 1>x>0. Put . Then we obtain
$\int \frac{1+{x}^{2}}{\left(1-{x}^{2}\right)\sqrt{1+{x}^{4}}}dx=\int \frac{1+y}{2\left(1-y\right)\sqrt{1+{y}^{2}}\sqrt{y}}dy$
Introduce the new variable

Then we have
$y=\frac{-1+t}{1+t}$
$y=\frac{2}{\left(1=t{\right)}^{2}}dt$
Substituting back we obtain
$\int \frac{1+y}{2\left(1-y\right)\sqrt{1+{y}^{2}}\sqrt{y}}dy=\int \frac{t}{2\sqrt{1+\left(\frac{-1+t}{1+t}{\right)}^{2}\sqrt{\frac{-1+t}{1+t}}}}\frac{2}{\left(1+t{\right)}^{2}}dt$
$=\frac{1}{\sqrt{2}}\int \frac{t}{\sqrt{{t}^{4}-1}}dt$
$=\frac{1}{2\sqrt{2}}\mathrm{ln}\left({t}^{2}+\sqrt{{t}^{4}-1}\right)+C$
Putting back everything we obtain
$\frac{1}{2\sqrt{2}}\mathrm{ln}\left(\frac{\left(1+{x}^{2}{\right)}^{2}+2\sqrt{2}x\sqrt{1+{x}^{4}}}{\left(1-{x}^{2}{\right)}^{2}}\right)+C$