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.
–
Vighnesh Shenoy
·
9 months ago

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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. – Vighnesh Shenoy · 9 months ago

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– Kalpa Roy · 9 months ago

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can anyone please help me out with this integration – Kalpa Roy · 9 months, 1 week ago

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Se puede factorar cimplentando trinomio cuadrado perfecto (x^2+1-raiz(2)x)(x^2+1+raiz(2)x) – Carlos Suarez · 9 months ago

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Hint: \(1+x^4 = (x^2 + i)(x^2- i ) \) for \(i = \sqrt{-1} \), apply partial fractions. – Pi Han Goh · 9 months, 1 week agoLog in to reply

– Kalpa Roy · 9 months, 1 week ago

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First application of integration of f(x)/g(x). Then application of integration of f (x)+g(x) In denominator – Arpan Manchanda · 9 months, 1 week ago

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