Respuesta :

LammettHash
One thing you could try is to set [tex]x=\sqrt y[/tex]. This makes [tex]y=x^2[/tex], so that [tex]\mathrm dy=2x\,\mathrm dx[/tex], and [tex]y^2=x^4[/tex]. So the integral is

[tex]\displaystyle\int\frac x{x^4+1}\,\mathrm dx=\frac12\int\frac{\mathrm dy}{y^2+1}=\frac12\arctan y+C=\frac12\arctan(x^2)+C[/tex]

A "trickier" way to do it is to write

[tex]x^4+1=(x^2+\sqrt2x+1)(x^2-\sqrt2x+1)[/tex]

so you could decompose the integrand into partial fractions. But that's more work than needed.

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