<s>[INST] 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.
 [/INST] 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. $C=C_1+C_2$.
</s>