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

Question

Cómo puedo solucionar integral de tipos

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

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

Answer 1

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

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

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

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

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

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

(2) debe ser tratada de la misma manera, poniendo de $x=\sqrt{\sec t}$

Answer 2

Se parece a la sustitución de variables en el Laboratorio de Bhattacharjee la respuesta puede ser generalizado para manejar indefinida de las integrales de la forma: $$ \int \frac{dx}{\sqrt[n]{x^{n}+1}}$$

Deje $y = \frac{x}{\sqrt[n]{x^n+1}}$, tenemos:

$$n y^{n-1} dy = \frac{n x^{n-1} dx}{(1+x^n)^2} \implies \frac{dy}{y} = \frac{dx}{x(1+x^n)} = ( 1 - y^n)\frac{dx}{x} \implies \frac{dx}{\sqrt[n]{x^n+1}} = \frac{dy}{1-y^n}$$

Deje $\zeta$ $n^{th}$ raíz primitiva de $1$, tenemos:

$$\int\frac{dy}{1 - y^n} = -\int\left[\frac{1}{n} \sum_{k=0}^{n-1} \frac{\zeta^k}{y - \zeta^k}\right]dy = -\frac{1}{n}\sum_{k=0}^{n-1}\zeta^k \log(1-\frac{y}{\zeta^k}) + C \tag{*}$$

Para $n = 4$, R. H. S de (*) se reduce a

$$\begin{align} &-\frac14 \left( \log(1-y) - \log(1+y) + i \left[\log(1-\frac{y}{i}) - \log(1 + \frac{y}{i})\right]\right) + C \\ =& \frac12 \left[ \frac12 \log(\frac{1+y}{1-y}) + \frac{1}{2i}\log(\frac{1+yi}{1-yi}) \right] + C\\ =& \frac12 \left[ \frac12 \log(\frac{1+y}{1-y}) + \arctan(y) \right] + C \end{align}$$

Para otros $n$, vamos a $c_k = \cos(\frac{2k\pi}{n})$$s_k = \sin(\frac{2k\pi}{n})$.

Al $n$ es, incluso, ( * ) se reduce a:

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

Al $n$ es impar, (*) se reduce a:

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

Answer 3

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

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

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

(1) debe ser tratada de la misma manera, poniendo de $1+\frac1{x^4}=y^4$

Answer 4

Ninguno de estos es de primaria. Maple y Mathematica expresar mediante funciones hipergeométricas.