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

Question Number 49829 by mhozhez last updated on 11/Dec/18

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

Commented by maxmathsup by imad last updated on 11/Dec/18

l e t I = \int \frac{x^{2}}{x^{4} + 1} d x v h a n g e m e n t x = \frac{1}{t} g i v e

I = \int \frac{1}{t^{2} (\frac{1}{t^{4}} + 1)} - \frac{d t}{t^{2}} = - \int \frac{d t}{t^{4} (1 + \frac{1}{t^{4}})} = \int \frac{d t}{t^{4} + 1} l e t d e c o m p o s e

F (t) = \frac{1}{1 + t^{4}} \Rightarrow F (t) = \frac{1}{{(t^{2} + 1)}^{2} - 2 t^{2}} = \frac{1}{(t^{2} + \sqrt{2} t + 1) (t^{2} - \sqrt{2} t + 1)}

= \frac{a t + b}{t^{2} + \sqrt{2} t + 1} + \frac{c t + d}{t^{2} - \sqrt{2} t + 1} w e h a v e F (- t) = F (t) \Rightarrow

\frac{- a t + b}{t^{2} - \sqrt{2} t + 1} + \frac{- c t + d}{t^{2} + \sqrt{2} t + 1} = F (t) \Rightarrow c = - a a n d b = d \Rightarrow

F (t) = \frac{a t + b}{t^{2} + \sqrt{2} t + 1} + \frac{- a t + b}{t^{2} - \sqrt{2} t + 1}

F (0) = 1 = 2 b \Rightarrow b = \frac{1}{2}

F (1) = \frac{1}{2} = \frac{a + b}{2 + \sqrt{2}} + \frac{- a + b}{2 - \sqrt{2}} \Rightarrow \frac{(2 - \sqrt{2}) (a + b) + (2 + \sqrt{2}) (- a + b)}{2} = \frac{1}{2} \Rightarrow

(2 - \sqrt{2} - 2 - \sqrt{2}) a + (2 - \sqrt{2} + 2 + \sqrt{2}) b = 1 \Rightarrow

- 2 \sqrt{2} a + 4 . \frac{1}{2} = 1 \Rightarrow - 2 \sqrt{2} a = - 1 \Rightarrow a = \frac{1}{2 \sqrt{2}} \Rightarrow

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

\int F (x) d x = \frac{1}{2 \sqrt{2}} \int \frac{t + \sqrt{2}}{t^{2} + \sqrt{2} t + 1} d t - \frac{1}{2 \sqrt{2}} \int \frac{t - \sqrt{2}}{t^{2} - \sqrt{2} t + 1} d t = H - K

2 \sqrt{2} H = \frac{1}{2} \int \frac{2 t + \sqrt{2} + \sqrt{2}}{t^{2} + \sqrt{2} t + 1} d t = \frac{1}{2} l n (t^{2} + \sqrt{2} t + 1) + \frac{\sqrt{2}}{2} \int \frac{d t}{t^{2} + \sqrt{2} t + 1}

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

=_{t + \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}} u} 2 \int \frac{1}{1 + u^{2}} \frac{d u}{\sqrt{2}} = \sqrt{2} a r c t a n (\frac{t \sqrt{2} + 1}{\sqrt{2}}) + c \Rightarrow

2 \sqrt{2} H = \frac{1}{2} l n (t^{2} + \sqrt{2} t + 1) + a r c t a n (\frac{t \sqrt{2} + 1}{\sqrt{2}}) + c b u t t = \frac{1}{x} \Rightarrow

H = \frac{1}{4 \sqrt{2}} l n (\frac{1}{x^{2}} + \frac{\sqrt{2}}{x} + 1) + \frac{1}{2 \sqrt{2}} a r c t a n (\frac{1}{x} + \frac{1}{\sqrt{2}}) + c w e f o l l o w t h e s a m e

m a n n e r f o r c a l c u l u s o f K .

Answered by tanmay.chaudhury50@gmail.com last updated on 11/Dec/18

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

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

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

= \frac{1}{2} \times \frac{1}{2 \sqrt{2}} l n {\frac{(x + \frac{1}{x}) - \sqrt{2}}{(x + \frac{1}{x}) + \sqrt{2}}} + \frac{1}{2} \times \frac{1}{\sqrt{2}} {t a n}^{- 1} (\frac{x - \frac{1}{x}}{\sqrt{2}}) + c

p l s c h e c k \dots

Commented by mhozhez last updated on 11/Dec/18

t h a n k s s i r

Commented by tanmay.chaudhury50@gmail.com last updated on 11/Dec/18

m o s t w e l c o m e \dots