Find-the-integral-dx-x-4-x-2-1-

Question Number 2103 by Yozzi last updated on 02/Nov/15

F i n d t h e i n t e g r a l

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

Commented by Yozzi last updated on 02/Nov/15

Y u p .

Commented by prakash jain last updated on 02/Nov/15

one option .

\frac{1}{x^{4} - x^{2} + 1} = \frac{1}{x^{4} - x^{2} + \frac{1}{4} + \frac{3}{4}}

= \frac{1}{{(x^{2} - \frac{1}{2})}^{2} - {(\frac{i \sqrt{3}}{2})}^{2}}

= (\frac{1}{i \sqrt{3}}) [\frac{1}{x^{2} - \frac{1}{2} - \frac{i \sqrt{3}}{2}} - \frac{1}{x^{2} - \frac{1}{2} + \frac{i \sqrt{3}}{2}}]

Commented by Yozzi last updated on 03/Nov/15

I s t h a t t h e o n l y w a y o n e c o u l d

s o l v e t h a t i n t e g r a l ?

Commented by prakash jain last updated on 03/Nov/15

x = \tan u

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

\frac{1}{\frac{\sin^{4} u}{\cos^{2} u} - \sin^{2} u + 1} = \frac{\cos^{2} u}{\sin^{4} u - \sin^{2} u \cos^{2} u + \cos^{2} u}

= \frac{{2 \cos}^{2} u}{2 - {(\sin 2 u)}^{2}}

This can be integrated without using i .

Answered by prakash jain last updated on 03/Nov/15

\frac{d}{d u} \sin 2 u = 2 (\cos 2 u) = 2 ({2 \cos}^{2} u - 1)

d y = 2 \cos 2 u d u =

\frac{{2 \cos}^{2} u}{2 - {(\sin 2 u)}^{2}} = \frac{1}{2} [\frac{2 ({2 \cos}^{2} u - 1)}{2 - {(\sin 2 u)}^{2}} + \frac{2}{2 - {(\sin 2 u)}^{2}}]

p a r t A w i l l i n e g r a t e t o

\ln (\sqrt{2} - \sin 2 u) - \ln (\sqrt{2} + \sin 2 u)

\frac{2}{2 - {(\sin 2 u)}^{2}} t o b e d o n e

Commented by prakash jain last updated on 03/Nov/15

\int \frac{d u}{2 - \sin 2 u} = \frac{1}{2 \sqrt{2}} \tan^{- 1} (\frac{\tan 2 x}{\sqrt{2}})