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

Question Number 181319 by a.lgnaoui last updated on 23/Nov/22

D e t e r m i n e r

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

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

Answered by floor(10²Eta[1]) last updated on 23/Nov/22

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

u = x^{2} \Rightarrow \frac{du}{2} = xdx

I = \frac{1}{2} \int \frac{du}{u^{2} + u + 1} = \frac{2}{3} \int \frac{du}{{(\frac{2 u}{\sqrt{3}} + \frac{1}{\sqrt{3}})}^{2} + 1}

t = \frac{2 u + 1}{\sqrt{3}} \Rightarrow dt = \frac{2}{\sqrt{3}} du

I = \frac{\sqrt{3}}{3} \int \frac{dt}{t^{2} + 1} = \frac{\sqrt{3}}{3} arctg (t) + C

\Rightarrow I = \frac{\sqrt{3}}{3} arctg (\frac{{2 x}^{2} + 1}{\sqrt{3}}) + C

Commented by a.lgnaoui last updated on 24/Nov/22

t h a n k s