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

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

b2sonicxh 2022-01-04 Answered

How do I integrate the following:

\int \frac{1 + x^{2}}{(1 - x^{2}) \sqrt{1 + x^{4}}} d x

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Stuart Rountree

Answered 2022-01-05 Author has 29 answers

\int \frac{1 + x^{2}}{(1 - x^{2}) \sqrt{1 + x^{4}}} d x

Let

u = x - \frac{1}{x}

Then

d u = (1 + \frac{1}{x^{2}}) d x

Now

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

Thus

\int \frac{1 + x^{2}}{(1 - x^{2}) \sqrt{1 + x^{4}}} d x

= - \int \frac{d u}{u \sqrt{u^{2} + 2}}

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Elaine Verrett

Answered 2022-01-06 Author has 41 answers

Somewhat inspired by Morons

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karton

Answered 2022-01-11 Author has 368 answers

Without loss of generality we may assume that 1>x>0. Put $x := \sqrt{y}, 1 > y > 0$ . Then we obtain
$\int \frac{1 + x^{2}}{(1 - x^{2}) \sqrt{1 + x^{4}}} d x = \int \frac{1 + y}{2 (1 - y) \sqrt{1 + y^{2}} \sqrt{y}} d y$
Introduce the new variable
$t := \frac{1 + y}{1 - y}, 1 < t < \infty$
Then we have
$y = \frac{- 1 + t}{1 + t}$
$y = \frac{2}{(1 = t)^{2}} d t$
Substituting back we obtain
$\int \frac{1 + y}{2 (1 - y) \sqrt{1 + y^{2}} \sqrt{y}} d y = \int \frac{t}{2 \sqrt{1 + (\frac{- 1 + t}{1 + t})^{2} \sqrt{\frac{- 1 + t}{1 + t}}}} \frac{2}{(1 + t)^{2}} d t$
$= \frac{1}{\sqrt{2}} \int \frac{t}{\sqrt{t^{4} - 1}} d t$
$= \frac{1}{2 \sqrt{2}} \ln (t^{2} + \sqrt{t^{4} - 1}) + C$
Putting back everything we obtain
$\frac{1}{2 \sqrt{2}} \ln (\frac{(1 + x^{2})^{2} + 2 \sqrt{2} x \sqrt{1 + x^{4}}}{(1 - x^{2})^{2}}) + C$