x-2-1-x-4-1-dx-

Question Number 86491 by john santu last updated on 29/Mar/20

\int \frac{x^{2} + 1}{x^{4} + 1} dx ?

Commented by john santu last updated on 29/Mar/20

dear prof mr mjs . what the super

easy method ?

Answered by som(math1967) last updated on 29/Mar/20

\int \frac{\frac{x^{2} + 1}{x^{2}}}{\frac{x^{4} + 1}{x^{2}}} d x

\int \frac{1 + \frac{1}{x^{2}}}{x^{2} + \frac{1}{x^{2}}} d x

\int \frac{d (x - \frac{1}{x})}{{(x - \frac{1}{x})}^{2} + {(\sqrt{2})}^{2}}

\frac{1}{\sqrt{2}} \tan^{- 1} (\frac{x - \frac{1}{x}}{\sqrt{2}}) + C

Commented by john santu last updated on 29/Mar/20

good answer

Commented by som(math1967) last updated on 29/Mar/20

T h a n k y o u s i r

Answered by MJS last updated on 29/Mar/20

other method :

\int \frac{x^{2} + 1}{x^{4} + 1} d x =

= \frac{1}{2} \int \frac{d x}{x^{2} - \sqrt{2} x + 1} + \frac{1}{2} \int \frac{d x}{x^{2} + \sqrt{2} x + 1} =

= \frac{\sqrt{2}}{2} (\arctan (\sqrt{2} x - 1) + \arctan (\sqrt{2} x + 1)) + C

Commented by john santu last updated on 29/Mar/20

ostrogradetski ?

Commented by MJS last updated on 29/Mar/20

no . just decomposing

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