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

Question

How can I solve integral of types

(1) $\displaystyle \int\dfrac{1}{\sqrt[4]{x^4+1}}dx$

(2) $\displaystyle \int\dfrac{1}{\sqrt[4]{x^4-1}}dx$

Answer 1

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:

$$ -\frac{1}{n} \log(1-y) - \frac{2}{n} \sum_{k=1}^{\frac{n-1}{2}} \left[ c_k \log( 1 + y^2 - 2 c_k y) - s_k \arctan(\frac{s_k y}{1 - c_k y})\right] + C $$

Answer 2

Neither of these is elementary. Maple and Mathematica express them using hypergeometric functions.

Answer 3

(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 $x=\sqrt{\sec t}$

Answer 4

(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 $1+\frac1{x^4}=y^4$