∫ (1/(x^4 +1)) dx


Question and Answers Forum

All Questions Topic List
Integration Questions
Previous in All Question Next in All Question
Previous in Integration Next in Integration
Question Number 36649 by rahul 19 last updated on 03/Jun/18

$\int \frac{1}{x^{4} + 1} d x$
Commented by rahul 19 last updated on 03/Jun/18

$\int \frac{\frac{1}{x^{2}}}{x^{2} + \frac{1}{x^{2}}} d x$ $\Rightarrow \int \frac{1 + \frac{1}{x^{2}} - 1}{x^{2} + \frac{1}{x^{2}}} d x$ $\Rightarrow \int \frac{1 + \frac{1}{x^{2}}}{{(x - \frac{1}{x})}^{2} + 2} d x - \int \frac{1}{{(x - \frac{1}{x})}^{2} + 2} d x$ $Continue . . . .$
Commented by rahul 19 last updated on 03/Jun/18

$By doing this way i am getting wrong$ $result! Can someone continue from$ $here and do check ?$
Commented by abdo mathsup last updated on 03/Jun/18
$the best way is to decompose the fraction F(x) =(1/(1+x^2 ))$
$t h e b e s t w a y i s t o d e c o m p o s e t h e f r a c t i o n$ $F (x) = \frac{1}{1 + x^{2}}$
Commented by abdo mathsup last updated on 04/Jun/18
$let decompose the ftaction F(x)= (1/(x^4 +1)) F(x)= (1/((x^2 +1)^2 −2x^2 )) =(1/((x^2 +1 +x(√2) )(x^2 +1 −x(√2)))) = ((ax +b)/(x^2 +x(√2) +1))) +((cx+d)/(x^2 −x(√2) +1)) F(−x)=F(x) ⇒ ((−ax +b)/(x^2 −x(√2) +1)) + ((−cx +d)/(x^2 +x(√2) +1)) =F(x)⇒ a=−c and b=d ⇒ F(x)= ((ax+b)/(x^2 +x(√2) +1)) +((−ax +b)/(x^2 −x(√2) +1)) F(0) =1 = b +b =2b ⇒b=(1/2) F(x) = (1/2) ((2ax +1)/(x^2 +(√2)+1)) +(1/2) ((−2ax +1)/(x^2 −x(√2) +1)) F(1) =(1/2) =(1/2){ ((2a+1)/(2+(√2))) +((−2a+1)/(2−(√2)))} ⇒ (2−(√2))(2a+1) +(2+(√2))(−2a+1)=2 2(2−(√2))a +2−(√2) −2(2+(√2))a +2+(√2) =2 −4(√2) a +4 =2 ⇒−4(√2)a =−2 ⇒ a = (1/(2(√2))) so F(x)= (1/2){ (((x/(√2)) +1)/(x^2 +(√2)x +1)) +((−(x/(√2)) +1)/(x^2 −x(√2) x+1))} =(1/(2(√2))){ ((x+(√2))/(x^2 +(√2)x +1)) +((−x +(√2))/(x^2 −x(√2) +1))} 2(√2) I = ∫ ((x +(√2))/(x^2 +(√2)x+1))dx − ∫ ((x−(√2))/(x^2 −x(√2) +1))dx but ∫ ((x+(√2))/(x^2 +(√2)x+1))dx = (1/2)∫ ((2x +(√2) +(√2))/(x^2 +(√2)x +1))dx =(1/2)ln(x^2 +(√2)x +1) +((√2)/2) ∫ (dx/(x^2 +(√2)x +1)) ∫ (dx/(x^2 +(√2)x +1)) = ∫ (dx/((x+((√2)/2))^2 +(3/4))) and we use the chang. x+((√2)/2) =((√3)/2) t ∫ (...)dx =(4/3) ∫ (1/(1+t^2 )) ((√3)/2) dt =((2(√3))/3) arctan(((2x+(√2))/(√3))) +c .....$
$l e t d e c o m p o s e t h e f t a c t i o n$ $F (x) = \frac{1}{x^{4} + 1}$ $F (x) = \frac{1}{{(x^{2} + 1)}^{2} - 2 x^{2}} = \frac{1}{(x^{2} + 1 + x \sqrt{2}) (x^{2} + 1 - x \sqrt{2})}$ $= \frac{a x + b}{x^{2} + x \sqrt{2} + 1)} + \frac{c x + d}{x^{2} - x \sqrt{2} + 1}$ $F (- x) = F (x) \Rightarrow$ $\frac{- a x + b}{x^{2} - x \sqrt{2} + 1} + \frac{- c x + d}{x^{2} + x \sqrt{2} + 1} = F (x) \Rightarrow$ $a = - c a n d b = d \Rightarrow$ $F (x) = \frac{a x + b}{x^{2} + x \sqrt{2} + 1} + \frac{- a x + b}{x^{2} - x \sqrt{2} + 1}$ $F (0) = 1 = b + b = 2 b \Rightarrow b = \frac{1}{2}$ $F (x) = \frac{1}{2} \frac{2 a x + 1}{x^{2} + \sqrt{2} + 1} + \frac{1}{2} \frac{- 2 a x + 1}{x^{2} - x \sqrt{2} + 1}$ $F (1) = \frac{1}{2} = \frac{1}{2} {\frac{2 a + 1}{2 + \sqrt{2}} + \frac{- 2 a + 1}{2 - \sqrt{2}}} \Rightarrow$ $(2 - \sqrt{2}) (2 a + 1) + (2 + \sqrt{2}) (- 2 a + 1) = 2$ $2 (2 - \sqrt{2}) a + 2 - \sqrt{2} - 2 (2 + \sqrt{2}) a + 2 + \sqrt{2} = 2$ $- 4 \sqrt{2} a + 4 = 2 \Rightarrow - 4 \sqrt{2} a = - 2 \Rightarrow a = \frac{1}{2 \sqrt{2}} s o$ $F (x) = \frac{1}{2} {\frac{\frac{x}{\sqrt{2}} + 1}{x^{2} + \sqrt{2} x + 1} + \frac{- \frac{x}{\sqrt{2}} + 1}{x^{2} - x \sqrt{2} x + 1}}$ $= \frac{1}{2 \sqrt{2}} {\frac{x + \sqrt{2}}{x^{2} + \sqrt{2} x + 1} + \frac{- x + \sqrt{2}}{x^{2} - x \sqrt{2} + 1}}$ $2 \sqrt{2} I = \int \frac{x + \sqrt{2}}{x^{2} + \sqrt{2} x + 1} d x - \int \frac{x - \sqrt{2}}{x^{2} - x \sqrt{2} + 1} d x$ $b u t$ $\int \frac{x + \sqrt{2}}{x^{2} + \sqrt{2} x + 1} d x = \frac{1}{2} \int \frac{2 x + \sqrt{2} + \sqrt{2}}{x^{2} + \sqrt{2} x + 1} d x$ $= \frac{1}{2} l n (x^{2} + \sqrt{2} x + 1) + \frac{\sqrt{2}}{2} \int \frac{d x}{x^{2} + \sqrt{2} x + 1}$ $\int \frac{d x}{x^{2} + \sqrt{2} x + 1} = \int \frac{d x}{{(x + \frac{\sqrt{2}}{2})}^{2} + \frac{3}{4}} a n d w e u s e$ $t h e c h a n g . x + \frac{\sqrt{2}}{2} = \frac{\sqrt{3}}{2} t$ $\int (. . .) d x = \frac{4}{3} \int \frac{1}{1 + t^{2}} \frac{\sqrt{3}}{2} d t$ $= \frac{2 \sqrt{3}}{3} a r c t a n (\frac{2 x + \sqrt{2}}{\sqrt{3}}) + c . . . . .$
	Answered by ajfour last updated on 03/Jun/18

	$= \frac{1}{2} \int \frac{1 - \frac{1}{x^{2}}}{{(x + \frac{1}{x})}^{2} - {(\sqrt{2})}^{2}} d x$ $+ \frac{1}{2} \int \frac{1 + \frac{1}{x^{2}}}{{(x - \frac{1}{x})}^{2} + {(\sqrt{2})}^{2}} d x$ $= \frac{1}{4 \sqrt{2}} \ln ∣ \frac{x + \frac{1}{x} - \sqrt{2}}{x + \frac{1}{x} + \sqrt{2}} ∣$ $+ \frac{1}{2 \sqrt{2}} \tan^{- 1} (\frac{x - \frac{1}{x} + \sqrt{2}}{\sqrt{2}}) + c .$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum