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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 ))

$$\mathrm{partial}\:\mathrm{fraction}\:\mathrm{with}\:\mathrm{detail}\:\mathrm{step}\:\mathrm{by}\:\mathrm{step}\:\mathrm{from} \\ $$$$\frac{\mathrm{4x}^{\mathrm{3}} +\mathrm{x}^{\mathrm{2}} +\mathrm{1}}{\left(\mathrm{x}^{\mathrm{2}} −\mathrm{2}\right)^{\mathrm{3}} } \\ $$

Answered by mathmax by abdo last updated on 14/Dec/20

F(x)=((4x^3 +x^2 +1)/((x^2 −2)^3 )) ⇒F(x)=((4x^3 +x^2 +1)/((x−(√2))^3 (x+(√2))^3 )) =(a/(x−(√2)))+(b/((x−(√2))^2 )) +(c/((x−(√2))^3 )) +(d/(x+(√2))) +(e/((x+(√2))^2 )) +(d/((x+(√2))^3 )) we can know c and d immediatly c=(x−2)^3 F(x)∣_(x=(√2)) =((4((√2))^3 +2+1)/((2(√2))^3 ))=((8(√2)+3)/(16(√2))) d =(x+(√2))F(x)∣_(x=−(√2)) =((4(−(√2))^3 +2+1)/((−2(√2))^3 )) =.... lim_(x→+∞) xF(x)=0 =a+d ⇒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...

$$\mathrm{F}\left(\mathrm{x}\right)=\frac{\mathrm{4x}^{\mathrm{3}} \:+\mathrm{x}^{\mathrm{2}} \:+\mathrm{1}}{\left(\mathrm{x}^{\mathrm{2}} −\mathrm{2}\right)^{\mathrm{3}} }\:\Rightarrow\mathrm{F}\left(\mathrm{x}\right)=\frac{\mathrm{4x}^{\mathrm{3}} \:+\mathrm{x}^{\mathrm{2}} \:+\mathrm{1}}{\left(\mathrm{x}−\sqrt{\mathrm{2}}\right)^{\mathrm{3}} \left(\mathrm{x}+\sqrt{\mathrm{2}}\right)^{\mathrm{3}} } \\ $$$$=\frac{\mathrm{a}}{\mathrm{x}−\sqrt{\mathrm{2}}}+\frac{\mathrm{b}}{\left(\mathrm{x}−\sqrt{\mathrm{2}}\right)^{\mathrm{2}} }\:+\frac{\mathrm{c}}{\left(\mathrm{x}−\sqrt{\mathrm{2}}\right)^{\mathrm{3}} }\:+\frac{\mathrm{d}}{\mathrm{x}+\sqrt{\mathrm{2}}}\:+\frac{\mathrm{e}}{\left(\mathrm{x}+\sqrt{\mathrm{2}}\right)^{\mathrm{2}} }\:+\frac{\mathrm{d}}{\left(\mathrm{x}+\sqrt{\mathrm{2}}\right)^{\mathrm{3}} } \\ $$$$\mathrm{we}\:\mathrm{can}\:\mathrm{know}\:\mathrm{c}\:\mathrm{and}\:\mathrm{d}\:\:\:\:\mathrm{immediatly} \\ $$$$\mathrm{c}=\left(\mathrm{x}−\mathrm{2}\right)^{\mathrm{3}} \:\mathrm{F}\left(\mathrm{x}\right)\mid_{\mathrm{x}=\sqrt{\mathrm{2}}} \:\:=\frac{\mathrm{4}\left(\sqrt{\mathrm{2}}\right)^{\mathrm{3}} \:+\mathrm{2}+\mathrm{1}}{\left(\mathrm{2}\sqrt{\mathrm{2}}\right)^{\mathrm{3}} }=\frac{\mathrm{8}\sqrt{\mathrm{2}}+\mathrm{3}}{\mathrm{16}\sqrt{\mathrm{2}}} \\ $$$$\mathrm{d}\:=\left(\mathrm{x}+\sqrt{\mathrm{2}}\right)\mathrm{F}\left(\mathrm{x}\right)\mid_{\mathrm{x}=−\sqrt{\mathrm{2}}} \:\:\:\:=\frac{\mathrm{4}\left(−\sqrt{\mathrm{2}}\right)^{\mathrm{3}} \:+\mathrm{2}+\mathrm{1}}{\left(−\mathrm{2}\sqrt{\mathrm{2}}\right)^{\mathrm{3}} }\:\:\:=.... \\ $$$$\boldsymbol{\mathrm{lim}}_{\boldsymbol{\mathrm{x}}\rightarrow+\infty} \mathrm{xF}\left(\mathrm{x}\right)=\mathrm{0}\:=\mathrm{a}+\mathrm{d}\:\Rightarrow\mathrm{d}=−\mathrm{a} \\ $$$$\mathrm{we}\:\mathrm{can}\:\mathrm{calculate}\:\mathrm{F}\left(\mathrm{0}\right)\:,\mathrm{F}\left(\mathrm{1}\right)\:\mathrm{and}\:\mathrm{F}\left(−\mathrm{1}\right)\mathrm{to}\:\mathrm{get}\:\mathrm{a}\:\mathrm{system}\:\mathrm{after} \\ $$$$\mathrm{we}\:\mathrm{slve}\:\mathrm{it}....\mathrm{any}\:\mathrm{way}\:\mathrm{there}\:\mathrm{is}\:\mathrm{a}\:\mathrm{lots}\:\mathrm{of}\:\mathrm{calculus}... \\ $$

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...

$$\mathrm{another}\:\mathrm{way}\:\mathrm{by}\:\mathrm{integral}\:\:\mathrm{we}\:\mathrm{have} \\ $$$$\mathrm{F}\left(\mathrm{x}\right)=\frac{\mathrm{d}}{\mathrm{dx}}\left(\int\:\mathrm{F}\left(\mathrm{x}\right)\mathrm{dx}\right)\:\:\mathrm{we}\:\mathrm{have}\:\int\:\mathrm{F}\left(\mathrm{x}\right)\mathrm{dx}=\int\:\frac{\mathrm{4x}^{\mathrm{3}} \:+\mathrm{x}^{\mathrm{2}} \:+\mathrm{1}}{\left(\mathrm{x}^{\mathrm{2}} −\mathrm{2}\right)^{\mathrm{3}} }\mathrm{dx} \\ $$$$=\int\:\:\frac{\mathrm{4x}^{\mathrm{3}} \:+\mathrm{x}^{\mathrm{2}} \:+\mathrm{1}}{\left(\mathrm{x}−\sqrt{\mathrm{2}}\right)^{\mathrm{3}} \left(\mathrm{x}+\sqrt{\mathrm{2}}\right)^{\mathrm{3}} }\mathrm{dx}\:=\int\:\:\:\frac{\mathrm{4x}^{\mathrm{3}} \:+\mathrm{x}^{\mathrm{2}} \:+\mathrm{1}}{\left(\frac{\mathrm{x}−\sqrt{\mathrm{2}}}{\mathrm{x}+\sqrt{\mathrm{2}}}\right)^{\mathrm{3}} \left(\mathrm{x}+\sqrt{\mathrm{2}}\right)^{\mathrm{6}} }\mathrm{dx} \\ $$$$\mathrm{we}\:\mathrm{do}\:\mathrm{the}\:\mathrm{changement}\:\frac{\mathrm{x}−\sqrt{\mathrm{2}}}{\mathrm{x}+\sqrt{\mathrm{2}}}=\mathrm{t}\:\Rightarrow\mathrm{x}−\sqrt{\mathrm{2}}=\mathrm{tx}+\sqrt{\mathrm{2}}\mathrm{t}\:\Rightarrow \\ $$$$\left(\mathrm{1}−\mathrm{t}\right)\mathrm{x}=\sqrt{\mathrm{2}}\mathrm{t}\:+\sqrt{\mathrm{2}}\:\Rightarrow\mathrm{x}=\frac{\sqrt{\mathrm{2}}\mathrm{t}+\sqrt{\mathrm{2}}}{\mathrm{1}−\mathrm{t}}\:\Rightarrow\frac{\mathrm{dx}}{\mathrm{dt}}=\frac{\sqrt{\mathrm{2}}\left(\mathrm{1}−\mathrm{t}\right)+\sqrt{\mathrm{2}}\mathrm{t}+\sqrt{\mathrm{2}}}{\left(\mathrm{1}−\mathrm{t}\right)^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{2}\sqrt{\mathrm{2}}}{\left(\mathrm{t}−\mathrm{1}\right)^{\mathrm{2}} }\:\:\mathrm{and}\:\mathrm{x}+\sqrt{\mathrm{2}}=\frac{\sqrt{\mathrm{2}}\mathrm{t}+\sqrt{\mathrm{2}}}{\mathrm{1}−\mathrm{t}}+\sqrt{\mathrm{2}}=\frac{\mathrm{2}\sqrt{\mathrm{2}}}{\mathrm{1}−\mathrm{t}}\:\Rightarrow \\ $$$$\int\:\mathrm{F}\left(\mathrm{x}\right)\mathrm{dx}\:=\int\:\:\:\frac{\mathrm{4}\left(\frac{\sqrt{\mathrm{2}}\mathrm{t}+\sqrt{\mathrm{2}}}{\mathrm{1}−\mathrm{t}}\right)^{\mathrm{3}} \:+\left(\frac{\sqrt{\mathrm{2}}\mathrm{t}+\sqrt{\mathrm{2}}}{\mathrm{1}−\mathrm{t}}\right)^{\mathrm{2}} \:+\mathrm{1}}{\mathrm{t}^{\mathrm{3}} \left(\frac{\mathrm{2}\sqrt{\mathrm{2}}}{\mathrm{1}−\mathrm{t}}\right)^{\mathrm{6}} }×\frac{\mathrm{2}\sqrt{\mathrm{2}}}{\left(\mathrm{t}−\mathrm{1}\right)^{\mathrm{2}} }\mathrm{dt} \\ $$$$=\frac{\mathrm{1}}{\left(\mathrm{2}\sqrt{\mathrm{2}}\right)^{\mathrm{5}} }\int\:\frac{\mathrm{1}}{\left(\mathrm{1}−\mathrm{t}\right)^{\mathrm{3}} }\left(\mathrm{4}\left(\sqrt{\mathrm{2}}\mathrm{t}+\sqrt{\mathrm{2}}\right)^{\mathrm{3}} \:+\left(\mathrm{1}−\mathrm{t}\right)\left(\sqrt{\mathrm{2}}\mathrm{t}+\sqrt{\mathrm{2}}\right)^{\mathrm{2}} +\left(\mathrm{1}−\mathrm{t}\right)^{\mathrm{3}} \right)\mathrm{2}\sqrt{\mathrm{2}}\left(\mathrm{1}−\mathrm{t}\right)^{\mathrm{4}} \:\:\frac{\mathrm{dt}}{\mathrm{t}^{\mathrm{3}} } \\ $$$$=−\frac{\mathrm{1}}{\left(\mathrm{2}\sqrt{\mathrm{2}}\right)^{\mathrm{5}} }\int\:\:\left(\mathrm{t}−\mathrm{1}\right)\left\{\mathrm{4}\left(\sqrt{\mathrm{2}}\mathrm{t}+\sqrt{\mathrm{2}}\right)^{\mathrm{3}} +\left(\mathrm{1}−\mathrm{t}\right)\left(\sqrt{\mathrm{2}}\mathrm{t}+\sqrt{\mathrm{2}}\right)^{\mathrm{2}} \:+\left(\mathrm{1}−\mathrm{t}\right)^{\mathrm{3}} \right)\mathrm{dx} \\ $$$$\mathrm{its}\:\mathrm{eazy}\:\mathrm{to}\:\mathrm{find}\:\mathrm{this}\:\mathrm{integral}\:\:\mathrm{so}\:\mathrm{you}\:\mathrm{get}\:\mathrm{decompositino}\:\mathrm{of} \\ $$$$\int\:\mathrm{F}\left(\mathrm{x}\right)\mathrm{dx}\:\:\mathrm{after}\:\mathrm{we}\:\mathrm{derivate}\:\mathrm{the}\:\mathrm{integral}\:\:\mathrm{this}\:\mathrm{method}\:\mathrm{is}\:\mathrm{sure}... \\ $$

Commented by joki last updated on 15/Dec/20

thank you sir for your solution.

$$ \\ $$$$\mathrm{thank}\:\mathrm{you}\:\mathrm{sir}\:\mathrm{for}\:\mathrm{your}\:\mathrm{solution}. \\ $$

Commented by mathmax by abdo last updated on 16/Dec/20

you are welcome

$$\mathrm{you}\:\mathrm{are}\:\mathrm{welcome}\: \\ $$

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