Refer-to-Q181319-x-4-1-x-4-x-2-dx-x-4-1-x-4-x-2-1-x-2-x-2-x-1-x-2-x-2-x-1-1-1-4-2x-1-1-x-2-x-1-1-4-2x-1-1-x-2-x-1-1-4-log-x-2-x-1-1-4-

Question Number 181403 by a.lgnaoui last updated on 24/Nov/22

R e f e r t o Q 181319

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

\frac{x^{4} + 1}{x^{4} + x + 2} = 1 + \frac{x}{2 (x^{2} + x + 1)} - \frac{x}{2 (x^{2} - x + 1)}

= 1 + \frac{1}{4} [\frac{(2 x + 1) - 1}{x^{2} + x + 1} - \frac{1}{4} \times \frac{(2 x - 1) + 1}{(x^{2} - x + 1)}

= \frac{1}{4} \times {[\log (x^{2} + x + 1)]}^{^{'}} - \frac{1}{4} \times (\frac{1}{[\frac{\sqrt{3}}{2} {(x + \frac{1}{2})}^{2}] + 1})

- (\frac{1}{4} {[\log (x^{2} - x + 1)]}^{^{'}} + \frac{1}{4} (\frac{1}{[\frac{\sqrt{3}}{2} \times {(x - \frac{1}{2}]}^{2} + 1}))

f i r s t w i t h [\frac{\sqrt{3}}{2} (x + \frac{1}{2})] = u \int \frac{1}{u^{2} + 1} d u = a r c t g (u)

a n d v = [\frac{\sqrt{3}}{2} (x - \frac{1}{2})] = v \int \frac{d v}{1 + v^{2}} = a r c t g (v)

\dots \dots \dots \dots \dots

Answered by ARUNG_Brandon_MBU last updated on 24/Nov/22

I = \int \frac{x^{4} + 1}{x^{4} + x^{2} + 1} d x = \int \frac{(x^{4} + x^{2} + 1) - x^{2}}{x^{4} + x^{2} + 1} d x

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

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

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

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

= x - \frac{1}{2} \int \frac{1 + \frac{1}{x^{2}}}{{(x - \frac{1}{x})}^{2} + 3} d x - \frac{1}{2} \int \frac{1 - \frac{1}{x^{2}}}{{(x + \frac{1}{x})}^{2} - 1} d x

= x - \frac{1}{2 \sqrt{3}} \arctan [\frac{1}{\sqrt{3}} (x - \frac{1}{x})] + \frac{1}{2} arcoth (x + \frac{1}{x}) + C

= x - \frac{\sqrt{3}}{6} \arctan (\frac{x^{2} - 1}{x \sqrt{3}}) + \frac{1}{4} \ln ∣ \frac{x^{2} + x + 1}{x^{2} - x + 1} ∣ + C