<s>[INST] Integral of type $\displaystyle \int\frac{1}{\sqrt[4]{x^4+1}}dx$ How can I solve integral of types
(1) $\displaystyle \int\dfrac{1}{\sqrt[4]{x^4+1}}dx$
(2) $\displaystyle \int\dfrac{1}{\sqrt[4]{x^4-1}}dx$
 [/INST] Looks like the variable substitution in Lab Bhattacharjee's answer can be generalized to handle indefinite integrals of the form:
$$ \int \frac{dx}{\sqrt[n]{x^{n}+1}}$$
Let $y = \frac{x}{\sqrt[n]{x^n+1}}$, we have: 
$$n y^{n-1} dy = \frac{n x^{n-1} dx}{(1+x^n)^2} \implies \frac{dy}{y} = \frac{dx}{x(1+x^n)} = ( 1 - y^n)\frac{dx}{x} \implies \frac{dx}{\sqrt[n]{x^n+1}} = \frac{dy}{1-y^n}$$
Let $\zeta$ be a $n^{th}$ primitive root of $1$, we have:
$$\int\frac{dy}{1 - y^n} = -\int\left[\frac{1}{n} \sum_{k=0}^{n-1} \frac{\zeta^k}{y - \zeta^k}\right]dy = -\frac{1}{n}\sum_{k=0}^{n-1}\zeta^k \log(1-\frac{y}{\zeta^k}) + C \tag{*}$$
For $n = 4$, R.H.S of (*) reduces to
$$\begin{align}
&-\frac14 \left( \log(1-y) - \log(1+y) + i \left[\log(1-\frac{y}{i}) - \log(1 + \frac{y}{i})\right]\right) + C \\
=&
\frac12 \left[ \frac12 \log(\frac{1+y}{1-y}) + \frac{1}{2i}\log(\frac{1+yi}{1-yi}) \right] + C\\
=&
\frac12 \left[ \frac12 \log(\frac{1+y}{1-y}) + \arctan(y) \right] + C
\end{align}$$
For other $n$, let $c_k = \cos(\frac{2k\pi}{n})$ and $s_k = \sin(\frac{2k\pi}{n})$.
When $n$ is even, (*) reduces to:
$$ \frac{1}{n} \log(\frac{1+y}{1-y}) - \frac{2}{n} \sum_{k=1}^{\frac{n}{2}-1}
\left[ c_k \log( 1 + y^2 - 2 c_k y) - s_k \arctan(\frac{s_k y}{1 - c_k y})\right] + C
$$
When $n$ is odd, (*) reduces to:
$$ -\frac{1}{n} \log(1-y) - \frac{2}{n} \sum_{k=1}^{\frac{n-1}{2}}
\left[ c_k \log( 1 + y^2 - 2 c_k y) - s_k \arctan(\frac{s_k y}{1 - c_k y})\right] + C
$$
</s>