Solve $$ \int\frac{dx}{(1+x^4)^{1/4}} $$
Set $t=\log x\implies x=e^t\implies dt=\dfrac{dx}{x}$ $$ \int\frac{dx}{(1+x^4)^{1/4}}=\int\frac{\dfrac{1}{x}dx}{\big(\dfrac{1}{x^4}+1\big)^{1/4}}=\int\frac{dt}{(e^{-4t}+1)^{1/4}} $$ Set $e^{-4t}+1=y\implies-4e^{-4t}dt=dy$ $$ I=\int\frac{e^{-4t}\,dt}{e^{-4t}(e^{-4t}+1)^{1/4}} = \frac{-1}{4}\int\frac{dy}{(y-1) y^{1/4}} $$
The solution given in my reference is $$\frac{1}{2} \left(\dfrac{1}{2}\log\dfrac{1+z}{1-z}-\tan^{-1}z\right),\quad z=\frac{(1+x^4)^{1/4}}{x}$$
How do I see what substitution to choose in order to find such a solution ?
Once you have $$I=\int\frac{dx}{(x-1)x^{1/4}},$$ take $x=u^4$ so that $dx=4u^3du$. We have $$I=4\int\frac{u^2du}{u^4-1}.$$ This is $$\begin{align} \tfrac14I&=\int\frac{u^2du}{(u^2+1)(u^2-1)}\\ &=\frac12\int\frac{du}{u^2+1}+\frac14\int\frac{du}{u-1}-\frac14\int\frac{du}{u+1}. \end{align}$$ The rest is easy from there.