x-x-4-1-dx-

Question Number 130287 by naka3546 last updated on 24/Jan/21

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

Answered by mathmax by abdo last updated on 24/Jan/21

I = \int \frac{xdx}{x^{4} + 1} let decompose F (x) = \frac{x}{x^{4} + 1} \Rightarrow

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

= \frac{ax + b}{x^{2} - \sqrt{2} x + 1} + \frac{mx + n}{x^{2} + \sqrt{2} x + 1} we see [that F (x) = \frac{1}{2 \sqrt{2}} (\frac{1}{x^{2} - \sqrt{2} x + 1} - \frac{1}{x^{2} + \sqrt{2} x + 1})

\Rightarrow \int F (x) dx = \frac{1}{2 \sqrt{2}} \int \frac{dx}{x^{2} - 2 \frac{\sqrt{2}}{2} x + \frac{1}{2} + \frac{1}{2}} - \frac{1}{2 \sqrt{2}} \int \frac{dx}{x^{2} + 2 \frac{\sqrt{2}}{2} x + \frac{1}{2} + \frac{1}{2}}

= \frac{1}{2 \sqrt{2}} \int \frac{dx}{{(x - \frac{1}{\sqrt{2}})}^{2} + \frac{1}{2}} (\to x - \frac{1}{\sqrt{2}} = \frac{u}{\sqrt{2}}) - \frac{1}{2 \sqrt{2}} \int \frac{dx}{{(x + \frac{1}{\sqrt{2}})}^{2} + \frac{1}{2}} (\to x + \frac{1}{\sqrt{2}} = \frac{v}{\sqrt{2}})

= \frac{1}{2 \sqrt{2}} \int \frac{du}{\sqrt{2} (\frac{1}{2} (u^{2} + 1))} - \frac{1}{2 \sqrt{2}} \int \frac{dv}{\sqrt{2} \frac{1}{2} (v^{2} + 1)}

= \frac{1}{2} \int \frac{du}{u^{2} + 1} - \frac{1}{2} \int \frac{dv}{v^{2} + 1} = \frac{1}{2} \arctan (u) - \frac{1}{2} arctanv + C

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

Commented by naka3546 last updated on 24/Jan/21

thanks, s i r .

H o w a b o u t \int \frac{x}{x^{4} + 3} d x ?

Commented by Ar Brandon last updated on 24/Jan/21

\int \frac{x}{x^{4} + 3} dx = \frac{1}{2} \int \frac{du}{u^{2} + 3} = \frac{\tan^{- 1} (x^{2} / \sqrt{3})}{2 \sqrt{3}} + C

Commented by mathmax by abdo last updated on 24/Jan/21

\int \frac{xdx}{x^{4} + 3} = \int \frac{xdx}{3 (\frac{x^{4}}{3} + 1)} = \int \frac{xdx}{3 ({(\frac{x}{(^{4} \sqrt{3})})}^{4} + 1)}

=_{\frac{x}{(^{4} \sqrt{3})} = t} \frac{1}{3} \int \frac{(^{4} \sqrt{3}) t}{1 + t^{4}} (^{4} \sqrt{3}) dt = \frac{3^{\frac{2}{4}}}{3} \int \frac{tdt}{1 + t^{4}} = \dots .

Commented by mathmax by abdo last updated on 26/Jan/21

you are welcome

Answered by Ar Brandon last updated on 24/Jan/21

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

= \frac{1}{2} \int \frac{du}{u^{2} + 1} = \frac{\tan^{- 1} (u)}{2} + C

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

Answered by mnjuly1970 last updated on 24/Jan/21

x^{2} = u

x d x = \frac{d u}{2}

ϕ = \int \frac{d u}{2 (u^{2} + 1)} = \frac{1}{2} {t a n}^{- 1} (x^{2}) + C \dots