Integral of type $\displaystyle \int\frac{1}{\sqrt[4]{x^4+1}}dx$

(1) Put $x=\sqrt{\tan t}\implies dx=\frac{\sec^2tdt}{2\sqrt{\tan t}}$ and $1+x^4=1+\tan^2t=\sec^2t$

So, $$\int\frac{dx}{(1+x^4)^\frac14}=\int\frac{\sec^2tdt}{2\sqrt{\tan t}\sqrt{\sec t}}=\int\frac{dt}{2\cos t\sqrt{\sin t}}=\int\frac{\cos tdt}{2\cos^2t\sqrt{\sin t}}$$

Put $\sin t=y^2\implies \cos tdt=2ydy $

$$\int\frac{\cos tdt}{2\cos^2t\sqrt{\sin t}}=\int\frac{2ydy}{2(1-y^4)y}=\frac12\left(\int\frac{dy}{1-y^2}+\int\frac{dy}{1+y^2}\right)$$

$$=\frac12\left(\frac12\ln\left|\frac{1+y}{1-y}\right|+\arctan y\right)+C$$

where $y^4=\sin^2t=\frac1{1+\cot^2t}=\frac1{1+\frac1{x^4}}=\frac{x^4}{1+x^4}$

(2) should be handled similarly by putting $x=\sqrt{\sec t}$

(2) $$\int \frac{dx}{(x^4-1)^\frac14}=\int\frac{dx}{x(1-\frac1{x^4})^\frac14}=\int\frac{x^4dx}{x^5(1-\frac1{x^4})^\frac14}$$

Let $1-\frac1{x^4}=y^4,4y^3dy=-4\frac{dx}{x^5}\implies \frac{dx}{x^5}=-y^3dy$ and $\frac1{x^4}=1-y^4\implies x^4=\frac1{1-y^4}$

$$\int\frac{x^4dx}{x^5(1-\frac1{x^4})^\frac14}$$ $$=\int\frac{-y^3dy}{(1-y^4)y}=\int\frac{y^2dy}{y^4-1}=\frac12\left(\frac{dy}{y^2-1}+\frac{dy}{1+y^2}\right)$$ $$=\frac12\left(\frac12\ln\left|\frac{y-1}{y+1}\right|+\arctan y\right)+C$$ where $y^4=1-\frac1{x^4}$

(1) should be handled similarly by putting $1+\frac1{x^4}=y^4$