partial fraction with detail step by step from ((4x^3 +x^2 +1)/((x^2 −2)^3 ))


Question and Answers Forum

All Questions Topic List
None Questions
Previous in All Question Next in All Question
Previous in None Next in None
Question Number 125867 by joki last updated on 14/Dec/20
$partial fraction with detail step by step from ((4x^3 +x^2 +1)/((x^2 −2)^3 ))$
$partial fraction with detail step by step from$ $\frac{{4 x}^{3} + x^{2} + 1}{{(x^{2} - 2)}^{3}}$
	Answered by mathmax by abdo last updated on 14/Dec/20

	$F (x) = \frac{{4 x}^{3} + x^{2} + 1}{{(x^{2} - 2)}^{3}} \Rightarrow F (x) = \frac{{4 x}^{3} + x^{2} + 1}{{(x - \sqrt{2})}^{3} {(x + \sqrt{2})}^{3}}$ $= \frac{a}{x - \sqrt{2}} + \frac{b}{{(x - \sqrt{2})}^{2}} + \frac{c}{{(x - \sqrt{2})}^{3}} + \frac{d}{x + \sqrt{2}} + \frac{e}{{(x + \sqrt{2})}^{2}} + \frac{d}{{(x + \sqrt{2})}^{3}}$ $we can know c and d immediatly$ $c = {(x - 2)}^{3} F (x) ∣_{x = \sqrt{2}} = \frac{4 {(\sqrt{2})}^{3} + 2 + 1}{{(2 \sqrt{2})}^{3}} = \frac{8 \sqrt{2} + 3}{16 \sqrt{2}}$ $d = (x + \sqrt{2}) F (x) ∣_{x = - \sqrt{2}} = \frac{4 {(- \sqrt{2})}^{3} + 2 + 1}{{(- 2 \sqrt{2})}^{3}} = . . . .$ $\lim_{x \to + \infty} xF (x) = 0 = a + d \Rightarrow d = - a$ $we can calculate F (0), F (1) and F (- 1) to get a system after$ $we slve it . . . . any way there is a lots of calculus . . .$
	Commented by mathmax by abdo last updated on 14/Dec/20
	$another way by integral we have F(x)=(d/dx)(∫ F(x)dx) we have ∫ F(x)dx=∫ ((4x^3 +x^2 +1)/((x^2 −2)^3 ))dx =∫ ((4x^3 +x^2 +1)/((x−(√2))^3 (x+(√2))^3 ))dx =∫ ((4x^3 +x^2 +1)/((((x−(√2))/(x+(√2))))^3 (x+(√2))^6 ))dx we do the changement ((x−(√2))/(x+(√2)))=t ⇒x−(√2)=tx+(√2)t ⇒ (1−t)x=(√2)t +(√2) ⇒x=(((√2)t+(√2))/(1−t)) ⇒(dx/dt)=(((√2)(1−t)+(√2)t+(√2))/((1−t)^2 )) =((2(√2))/((t−1)^2 )) and x+(√2)=(((√2)t+(√2))/(1−t))+(√2)=((2(√2))/(1−t)) ⇒ ∫ F(x)dx =∫ ((4((((√2)t+(√2))/(1−t)))^3 +((((√2)t+(√2))/(1−t)))^2 +1)/(t^3 (((2(√2))/(1−t)))^6 ))×((2(√2))/((t−1)^2 ))dt =(1/((2(√2))^5 ))∫ (1/((1−t)^3 ))(4((√2)t+(√2))^3 +(1−t)((√2)t+(√2))^2 +(1−t)^3 )2(√2)(1−t)^4 (dt/t^3 ) =−(1/((2(√2))^5 ))∫ (t−1){4((√2)t+(√2))^3 +(1−t)((√2)t+(√2))^2 +(1−t)^3 )dx its eazy to find this integral so you get decompositino of ∫ F(x)dx after we derivate the integral this method is sure...$
	$another way by integral we have$ $F (x) = \frac{d}{dx} (\int F (x) dx) we have \int F (x) dx = \int \frac{{4 x}^{3} + x^{2} + 1}{{(x^{2} - 2)}^{3}} dx$ $= \int \frac{{4 x}^{3} + x^{2} + 1}{{(x - \sqrt{2})}^{3} {(x + \sqrt{2})}^{3}} dx = \int \frac{{4 x}^{3} + x^{2} + 1}{{(\frac{x - \sqrt{2}}{x + \sqrt{2}})}^{3} {(x + \sqrt{2})}^{6}} dx$ $we do the changement \frac{x - \sqrt{2}}{x + \sqrt{2}} = t \Rightarrow x - \sqrt{2} = tx + \sqrt{2} t \Rightarrow$ $(1 - t) x = \sqrt{2} t + \sqrt{2} \Rightarrow x = \frac{\sqrt{2} t + \sqrt{2}}{1 - t} \Rightarrow \frac{dx}{dt} = \frac{\sqrt{2} (1 - t) + \sqrt{2} t + \sqrt{2}}{{(1 - t)}^{2}}$ $= \frac{2 \sqrt{2}}{{(t - 1)}^{2}} and x + \sqrt{2} = \frac{\sqrt{2} t + \sqrt{2}}{1 - t} + \sqrt{2} = \frac{2 \sqrt{2}}{1 - t} \Rightarrow$ $\int F (x) dx = \int \frac{4 {(\frac{\sqrt{2} t + \sqrt{2}}{1 - t})}^{3} + {(\frac{\sqrt{2} t + \sqrt{2}}{1 - t})}^{2} + 1}{t^{3} {(\frac{2 \sqrt{2}}{1 - t})}^{6}} \times \frac{2 \sqrt{2}}{{(t - 1)}^{2}} dt$ $= \frac{1}{{(2 \sqrt{2})}^{5}} \int \frac{1}{{(1 - t)}^{3}} (4 {(\sqrt{2} t + \sqrt{2})}^{3} + (1 - t) {(\sqrt{2} t + \sqrt{2})}^{2} + {(1 - t)}^{3}) 2 \sqrt{2} {(1 - t)}^{4} \frac{dt}{t^{3}}$ $= - \frac{1}{{(2 \sqrt{2})}^{5}} \int (t - 1) {4 {(\sqrt{2} t + \sqrt{2})}^{3} + (1 - t) {(\sqrt{2} t + \sqrt{2})}^{2} + {(1 - t)}^{3}) dx$ $its eazy to find this integral so you get decompositino of$ $\int F (x) dx after we derivate the integral this method is sure . . .$
	Commented by joki last updated on 15/Dec/20

	$thank you sir for your solution .$
	Commented by mathmax by abdo last updated on 16/Dec/20

	$you are welcome$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum