While trying to solve $\int\frac1{\sqrt[4]{1+x^4}}dx$, I got curious and tried substituting other integer values instead of 4. In other words, the following family of integrals:
$$\int\frac{1}{\sqrt[n]{x^n+1}}\,dx\qquad,\qquad n\in\mathbb{Z}_{\geq0}$$
eg:
$$\int\frac{1}{x+1}=\log(x+1)$$
$$\int\frac{1}{\sqrt{x^2+1}}=\log\left(\left|\sqrt{x^2+1}+x\right|\right)$$
$$\int\frac{1}{\sqrt[3]{x^3+1}}=\dfrac{\ln\left(\left|\left(x^3+1\right)^\frac{2}{3}+x\sqrt[3]{x^3+1}+x^2\right|\right)-2\ln\left(\left|\sqrt[3]{x^3+1}-x\right|\right)-2\sqrt{3}\arctan\left(\frac{2\sqrt[3]{x^3+1}+x}{\sqrt{3}x}\right)}{6}$$
$$\int\frac{1}{\sqrt[4]{x^4+1}}\,dx=\dfrac{\ln\left(\left|\sqrt[4]{x^4+1}+x\right|\right)-\ln\left(\left|\sqrt[4]{x^4+1}-x\right|\right)-2\arctan\left(\frac{\sqrt[4]{x^4+1}}{x}\right)}{4}$$
Just plugging them into integral-calculator.com, it would seem the antiderivative will always be expressable in elementary form as long as the root power (?) is equal to the exponent in the surd. At least up until $n=12$ before the calculator would time out. The expressions quickly become much more complicated from this point on.
However unlikely this may be, I was curious whether it would be possible to derive a general expression for the antiderivatives of this family of integrals. I must say I'm out of my depth at this point, but I just feel like this is an interesting idea and would like to put it out there for the community.
Best Answer
Continuing from bjorn93's answer, we can rewrite the integrand as $\frac{1}{t^{2}\left(t^{n}-1\right)}+\frac{1}{t^{2}}$, such that $n$ only appears once. We can then express this with partial fractions, as the denominator can be factored into the complex roots of unity, $\zeta^k$. Using Question 4803, we have $\displaystyle\frac{1}{\left(t^{n}-1\right)}=\sum_{k=0}^{n-1}\frac{\frac{\zeta^k}{n}}{t-\zeta^k}$. The integral can then be evaluated by swapping the integral and sum operators and using $\displaystyle \int \frac{1}{t^2(at-1)}\,\mathrm{d}t=-a\log t +a\log(1-at)+\frac1t+C$. Therefore, it is the real part of a finite sum of logarithms,
$$-\frac1n\int\sum_{k=0}^{n-1}\frac{1}{t^{2}\left(\frac1{\zeta^k}-1\right)}\ \mathrm{d}t-\int\frac{1}{t^{2}}\ \mathrm{d}t \\ =\frac{1}{n}\sum_{k=0}^{n-1}\frac{\operatorname{Log}t-\operatorname{Log}\left(t-\zeta^{k}\right)}{\zeta^{k}}+C $$
where $\operatorname{Log}(x)$ is the principal branch of the complex logarithm. We can then use $\operatorname{Log}(x)=\ln|x|+i\operatorname{Arg}(x)$, $\operatorname{Arg}(x)=\tan^{-1}\frac{\Im(x)}{\Re(x)}$ and $\zeta^k=c+is$, where $c=\cos\frac{2\pi k}{n}$ and $s=\sin\frac{2\pi k}{n}$ to show that, for $x\ge0$, the integral equals
$$\frac{1}{n}\sum_{k=0}^{n-1}\left(c\ln\frac{t}{\sqrt{\left(t-c\right)^{2}+s^{2}}}+s\tan^{-1}\left(\frac{s}{t-c}\right)\right)+C $$
which numerically agrees with the integral (Desmos link).