solutions to the family of integrals $\int\frac{1}{\sqrt[n]{x^n+1}}\,dx$

Question

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.

Answer 1

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).

Answer 2

The integral is, indeed, elementary. I'm going to assume $x\geq 0$. The case $x\leq 0$ is similar except for a minus sign when $n$ is even. Write the integrand as $$\frac{1}{\sqrt[n]{x^n+1}}=\left(x^n+1\right)^{-\frac 1n}=x^{-1}\left(1+x^{-n}\right)^{-\frac 1n} $$ and substitute $1+x^{-n}=t^n$. Then $-nx^{-n-1}\,dx=nt^{n-1}\,dt\Rightarrow -x^{-n}x^{-1}\,dx=t^{n-1}\,dt$ so $$x^{-1}\,dx=-\frac{t^{n-1}}{x^{-n}}\,dt=-\frac{t^{n-1}}{t^n-1}\,dt $$ Hence the integral becomes $$-\int\frac{t^{n-1}}{t^n-1}t^{-1}\,dt=-\int\frac{t^{n-2}}{t^n-1}\,dt $$ which is an integral of a rational function. We know it is elementary, and the answer only involves rational functions, logarithms, and arctangents. I don't have a closed-form formula for this integral but perhaps there is a recursive formula.