$\int\dfrac{dx}{x^2(x^4+1)^{3/4}}$

Question

Evaluar$$\large{\int\dfrac{dx}{x^2(x^4+1)^{3/4}}}$ $

Pensé en reescribir esto como$$\large{\int\dfrac{dx}{x^5(1+\frac{1}{x^4})^{3/4}}}$$ and substituting $$u^4=\left(1+\frac{1}{x^4}\right)\Rightarrow u=\left(1+\frac{1}{x^4}\right)^{1/4}$$ and subsequently I got $$du=\dfrac{1}{4}\left(1+\frac{1}{x^4}\right)^{-3/4}\times (-4x^{-5})dx$ $

Sin embargo, no puedo pensar cómo proceder más. Cualquier ayuda sería realmente apreciada. ¡Muchas gracias de antemano!

Answer 1

Tenemos $$ \ int \ frac {1} {x ^ {2} \ left (x ^ {4} +1 \ right) ^ {3/4}} dx \ overset {u = 1 / x ^ {4} } {=} - \ frac {1} {4} \ int \ frac {1} {\ left (u + 1 \ right) ^ {3/4}} du \ overset {u + 1 = v} {=} - \ frac {1} {4} \ int \ frac {1} {v ^ {3/4}} d =$$ $ $ = - \ sqrt [4] {v} + C = - \ frac { \ sqrt [4] {x ^ {4} +1}} {x} + C. $$

Answer 2

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La integral en la variable$$\int\frac{1}{x^5}\frac{1}{(1+\frac{1}{x^4})^{3/4}}dx$ es entonces

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Answer 3

Simplemente retomando donde lo dejaste:

$$ - \ int du \, = -u + C = - (1+ \ frac {1} {x ^ 4}) ^ {\ frac {1} {4}} + C. $$

Eso debería ser.

Answer 4

Dejar, $1+\frac{1}{x^4}=t \implies \frac{-4dx}{x^5}=dt$$$\large{\int\dfrac{dx}{x^2(x^4+1)^{3/4}}}$$ $$=\int \frac{dx}{x^5\left(1+\frac{1}{x^4}\right)^{3/4}}$$ $$=\frac{-1}{4}\int\frac{dt}{\left(t\right)^{3/4}}$$ $$=\frac{-1}{4}\int (t)^{-3/4}dt$$ $$=\frac{-1}{4}\frac{t^{1/4}}{1/4}+C$$ $$=-\left(1+\frac{1}{x^4}\right)^{1/4}+C$$$$ = - \ frac {(1 + x ^ 4) ^ {1/4}} {x} + C $$

Answer 5

$\bf{My\; Solution::}$ Let$\displaystyle I = \int\frac{1}{x^2\left(x^4+1\right)^{\frac{3}{4}}}dx\;,$ Let Let

Let$\displaystyle x = \frac{1}{t}\;,$ Then$\displaystyle dx = -\frac{1}{t^2}dt\;,$ Then Integral Convert into

$\displaystyle = -\int \frac{t^3}{(1+t^4)^{\frac{3}{4}}}dt\;,$ Ahora deja$(1+t^4) = u\;, $ Then$\displaystyle t^3dt = \frac{1}{4}du$

Entonces integral$\displaystyle = -\frac{1}{4}\int t^{-\frac{3}{4}}dt = -u^{\frac{1}{4}}+\mathcal{C} = -\left(1+t^4\right)^{\frac{1}{4}}+\mathcal{C}$

Entonces integral$\displaystyle \int\frac{1}{x^2\left(x^4+1\right)^{\frac{3}{4}}}dx = - \left(\frac{1+x^4}{x^4}\right)^{\frac{1}{4}}+\mathcal{C.}$