How to find this indefinite integral? $\int\frac{1+x^4}{(1-x^4)\cdot \sqrt{1+x^4}}dx$ I am thinking of a trig sub of $x^2 = \tan{t}$ but its not leading to a nice trigonmetric form, which i can integrate. Our teacher said that it can be computed using elementary methods, but I'm unable to think of the manipualtion.
 A: With parts of the calculation left as an exercise:
The surd-in-the-denominator expression I commented on is actually a subtle hint. Rewriting the numerator as$$1+x^4=\frac12(1-x^2)^2+\frac12(1+x^2)^2,$$we want to separately evaluate$$\frac12\int\frac{(1-x^2)^2}{(1-x^4)\sqrt{1+x^4}}dx=\frac12\int\frac{(1-x^2)}{(1+x^2)\sqrt{1+x^4}}dx$$and$$\frac12\int\frac{(1+x^2)^2}{(1-x^4)\sqrt{1+x^4}}dx=\frac12\int\frac{(1+x^2)}{(1-x^2)\sqrt{1+x^4}}dx.$$Write the first part as$$\frac{1}{2\sqrt{2}}\int\frac{dy}{\sqrt{1-y^2}}=\frac{1}{2\sqrt{2}}\arcsin y+C_1$$with $y:=\frac{x\sqrt{2}}{1+x^2}$. Similarly, with $z:=\frac{x\sqrt{2}}{\sqrt{1+x^4}}$ the second part is$$\frac{1}{2\sqrt{2}}\int\frac{dz}{1-z^2}=\frac{1}{2\sqrt{2}}\operatorname{artanh}z+C_2.$$ Adding these, subsume the integration constants into one viz. $C=C_1+C_2$.
A: \begin{aligned}\int{\frac{1+x^{4}}{\left(1-x^{4}\right)\sqrt{1+x^{4}}}\,\mathrm{d}x}&=\int{\frac{1+x^{4}}{\left(1-x^{2}\right)\left(1+x^{2}\right)\sqrt{1+x^{4}}}\,\mathrm{d}x}\\ &=\int{\frac{x^{2}+\frac{1}{x^{2}}}{\left(1-x^{2}\right)\left(x+\frac{1}{x}\right)\sqrt{x^{2}+\frac{1}{x^{2}}}}\,\mathrm{d}x}\\ &=\int{\frac{x^{2}+\frac{1}{x^{2}}}{\left(x-\frac{1}{x}\right)^{2}\left(x+\frac{1}{x}\right)\sqrt{x^{2}+\frac{1}{x^{2}}}}\left(\frac{1}{x^{2}}-1\right)\mathrm{d}x}\\ &=\int{\frac{\left(x+\frac{1}{x}\right)^{2}-2}{\left(\left(x+\frac{1}{x}\right)^{2}-4\right)\left(x+\frac{1}{x}\right)\sqrt{\left(x+\frac{1}{x}\right)^{2}-2}}\left(1-\frac{1}{x^{2}}\right)\mathrm{d}x}\\ &=\int{\frac{y^{2}-2}{y\left(y^{2}-4\right)\sqrt{y^{2}-2}}\,\mathrm{d}y}\end{aligned}
Using another substitution $ y = \sqrt{2}\cosh{t} $. We're left with the integral : $$ \frac{1}{2}\int{\frac{\tanh{t}}{\sinh^{2}{t}-1}\,\mathrm{d}t} $$
I'll let you take it from here.
Of caurse we made the assumption that we're working in a domain $ \subset \left]0,+\infty\right[ $, so we could have $ \sqrt{x^{2}}=x $. But the same approach will stand if we're working on a domain of $ \left]-\infty, 0\right[ $.
