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# Integration

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

Note by Kalpa Roy
1 year, 4 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. · 1 year, 4 months ago

thanks · 1 year, 4 months ago

Se puede factorar cimplentando trinomio cuadrado perfecto (x^2+1-raiz(2)x)(x^2+1+raiz(2)x) · 1 year, 4 months ago

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

thanx · 1 year, 4 months ago