How to evaluate the indefinite integral $\int\frac{1+x^2}{(1-x^2)(\sqrt 1+x^4)} dx$ in simplest form?

Question

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

I tried trigonometric substitution and partial fraction decomposition but it got me nowhere...

Answer 1

$$I=\int \frac{(1+x^2)}{(1-x^2)\sqrt{1+x^4}} dx=\int \frac{1+1/x^2}{(1/x-x)\sqrt{x^2+1/x^2}}=-\int \frac{1+1/x^2}{(x-1/x)\sqrt{(x-1/x)^2+4}}=$$ Let $x-1/x=t \implies (1+1/x^2)dx=dt,$ we get $$I=-\int\frac{dt}{t\sqrt{4+t^2}}=-\int \frac{2 sec^2 u}{2 \tan u ~2 sec u} u=\frac{-1}{2} \int \csc u du =-\frac{1}{2} \ln (u/2)+C,$$ Here $$u=2 \tan (x-1/x)$$ So $$I=\frac{-1}{2}\ln \tan (x-1/x)+C$$

Answer 2

First we rewrite our integral into a form that motivates the u-substitution $u=\frac{1}{x}-x$, thus:

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

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

Now we can perform the intended u-subtitution:

$$u=\frac{1}{x}-x$$ $$du=-(\frac{1}{x^2}+1)\,dx$$

Rewriting our integral in terms of $u$ gives the following:

$$-\int\frac{1}{u\sqrt{u^2+2}}\,du$$

Lastly we need to solve this integral which can be done in a mostly straightforward u-substitution, thus:

$$u=\sqrt{2}\tan{\theta}$$ $$du=\sqrt{2}\sec^2{\theta}\,d{\theta}$$

$$-\int\frac{\sqrt{2}\sec^2{\theta}}{\sqrt{2}\tan{\theta}\sqrt{2}\sec{\theta}}\,d{\theta}=-\frac{\sqrt{2}}{2}\int\csc{\theta}\,d{\theta}$$

$$-\frac{\sqrt{2}}{2}\int\csc{\theta}\,d{\theta}=\frac{\sqrt{2}}{2}\ln\vert{\csc{\theta}+\cot{\theta}}\vert+C$$

Now plugging back in our substitutions we have the following:

$$\frac{\sqrt{2}}{2}\ln\vert{\csc{\theta}+\cot{\theta}}\vert+C=\frac{\sqrt{2}}{2}\ln\left|{\frac{\sqrt{u^2+2}}{u}+\frac{\sqrt{2}}{u}}\right|+C$$

$$=\frac{\sqrt{2}}{2}\ln\left|{\frac{\sqrt{(\frac{1}{x}-x)^2+2}}{\frac{1}{x}-x}+\frac{\sqrt{2}}{\frac{1}{x}-x}}\right|+C=\frac{\sqrt{2}}{2}\ln\left|{\frac{\sqrt{1+x^4}+x\sqrt{2}}{1-x}}\right|+C$$

Therefore,

$$I=\int\frac{1+x^2}{(1-x^2)\sqrt{1+x^4}}\,dx=\frac{\sqrt{2}}{2}\ln\left|{\frac{\sqrt{1+x^4}+x\sqrt{2}}{1-x}}\right|+C$$