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\[ \large \int \dfrac1{1+x^4} \, dx = \, ? \]

Note by Kalpa Roy 1 year, 11 months ago

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A solution without complex numbers, \( I = \displaystyle \int \dfrac{1}{1+x^{4}}dx = \dfrac{1}{2} \int \dfrac{1+x^{2}}{1+x^{4}}dx - \dfrac{x^{2}-1}{1+x^{4}}dx \)

\( \therefore I = \dfrac{1}{2} \displaystyle \int \dfrac{1+\frac{1}{x^{2}}}{\left(x-\frac{1}{x}\right)^{2}+2}dx - \dfrac{1-\frac{1}{x^{2}}}{\left(x+\frac{1}{x}\right)^{2}-2}dx \)

In the first integral, substitute \( x - \dfrac{1}{x} = u \) and in the second integral substitute \( x + \dfrac{1}{x} = v \)

\( \therefore I = \dfrac{1}{2} \displaystyle \int \dfrac{du}{u^{2}+2} - \int \dfrac{dv}{v^{2}-2} \) I think you can continue after this.

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thanks

can anyone please help me out with this integration

Se puede factorar cimplentando trinomio cuadrado perfecto (x^2+1-raiz(2)x)(x^2+1+raiz(2)x)

Hint: \(1+x^4 = (x^2 + i)(x^2- i ) \) for \(i = \sqrt{-1} \), apply partial fractions.

thanx

First application of integration of f(x)/g(x). Then application of integration of f (x)+g(x) In denominator

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TopNewestA solution without complex numbers,

\( I = \displaystyle \int \dfrac{1}{1+x^{4}}dx = \dfrac{1}{2} \int \dfrac{1+x^{2}}{1+x^{4}}dx - \dfrac{x^{2}-1}{1+x^{4}}dx \)

\( \therefore I = \dfrac{1}{2} \displaystyle \int \dfrac{1+\frac{1}{x^{2}}}{\left(x-\frac{1}{x}\right)^{2}+2}dx - \dfrac{1-\frac{1}{x^{2}}}{\left(x+\frac{1}{x}\right)^{2}-2}dx \)

In the first integral, substitute \( x - \dfrac{1}{x} = u \) and in the second integral substitute \( x + \dfrac{1}{x} = v \)

\( \therefore I = \dfrac{1}{2} \displaystyle \int \dfrac{du}{u^{2}+2} - \int \dfrac{dv}{v^{2}-2} \)

I think you can continue after this.

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thanks

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can anyone please help me out with this integration

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Se puede factorar cimplentando trinomio cuadrado perfecto (x^2+1-raiz(2)x)(x^2+1+raiz(2)x)

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Hint: \(1+x^4 = (x^2 + i)(x^2- i ) \) for \(i = \sqrt{-1} \), apply partial fractions.Log in to reply

thanx

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First application of integration of f(x)/g(x). Then application of integration of f (x)+g(x) In denominator

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