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Question Number 35245 by JOHNMASANJA last updated on 17/May/18

$$express\frac{7x+4}{x^3+x^2+9x+9}inpartial$$$$fraction$$

Commented by Rasheed.Sindhi last updated on 17/May/18

$$\mathrm{prof}\mathrm{Abdo}\mathrm{imad}$$$$\mathrm{Use}\mathrm{of}\mathrm{limit}\mathrm{in}\mathrm{partial}\mathrm{fractions}\mathrm{is}\mathrm{new}$$$$\mathrm{for}\mathrm{me}.\mathrm{Could}\mathrm{you}\mathrm{please}\mathrm{insert}\mathrm{more}$$$$\mathrm{steps}\mathrm{to}\mathrm{discribe}4\mathrm{th},5\mathrm{th}\mathrm{\&}6\mathrm{th}\mathrm{lines}?$$

Commented by Rasheed.Sindhi last updated on 17/May/18

$$\mathrm{Sir}\mathrm{th}\propto \mathrm{n}\mathbb{k}\mathrm{\$}\mathrm{a}\lfloor \mathrm{♡}\mathrm{\top }!$$$$\mathrm{BTW}$$$$\mathrm{prof}\mathrm{Abdo}\mathrm{imad}\stackrel{?}{=}\mathrm{abdo}\mathrm{mathsup}649\mathrm{cc}$$$$\stackrel{?}{=}\mathrm{math}\mathrm{khazana}\mathrm{by}\mathrm{abdo}$$$$\stackrel{?}{=}\mathrm{abdo}.\mathrm{msup}.\mathrm{com}$$$$\stackrel{?}{=}\mathrm{ID}\mathrm{containing}{}^{\prime }{\mathrm{abdo}}^{\prime }$$$$???$$

Commented by prof Abdo imad last updated on 17/May/18

$$letputF\left(x\right)=\frac{7x+4}{x^3+x^2+9x+9}$$$$F\left(x\right)=\frac{7x+4}{x^2(x+1)+9(x+1)}=\frac{7x+4}{(x+1)(x^2+9)}$$$$F\left(x\right)=\frac{a}{x+1}+\frac{bx+c}{x^2+9}$$$$a={lim}_{x\to -1}(x+1)F\left(x\right)=\frac{-3}{10}$$$${lim}_{x\to +\mathrm{\infty }}xF\left(x\right)=0=a+b\Rightarrow b=\frac{3}{10}\Rightarrow$$$$F\left(x\right)=\frac{-3}{10(x+1)}+\frac{\frac{3}{10}x+c}{x^2+9}$$$$F\left(0\right)=\frac{4}{9}=\frac{-3}{10}+\frac{c}{9}\Rightarrow 4=-\frac{27}{10}+c\Rightarrow$$$$c=4+\frac{27}{10}=\frac{67}{10}\Rightarrow$$$$F\left(x\right)=\frac{-3}{10(x+1)}+\frac{1}{10}\frac{3x+67}{x^2+9}.$$

Commented by Rasheed.Sindhi last updated on 17/May/18

$$x^3+x^2+9x+9=x^2(x+1)+9(x+1)$$$$=(x+1)(x^2+9)$$$$\frac{7x+4}{(x+1)(x^2+9)}=\frac{\mathrm{A}}{x+1}+\frac{\mathrm{B}x+\mathrm{C}}{x^2+9}$$$$7x+4=\mathrm{A}(x^2+9)+(\mathrm{B}x+\mathrm{C})(x+1)$$$$x=-1\Rightarrow -3=10\mathrm{A}\Rightarrow \mathrm{A}=-\frac{3}{10}$$$$x=\pm 3i\Rightarrow 4\pm 21i=(\mathrm{B}(\pm 3i)+\mathrm{C})(\pm 3i+1)$$$$4\pm 21i=-9\mathrm{B}\pm 3i\mathrm{B}\pm 3i\mathrm{C}+\mathrm{C}$$$$=-9\mathrm{B}+\mathrm{C}\pm 3i(\mathrm{B}+\mathrm{C})$$$$-9\mathrm{B}+\mathrm{C}=4\wedge \mathrm{B}+\mathrm{C}=7$$$$\Rightarrow \mathrm{B}=\frac{3}{10}\wedge \mathrm{C}=\frac{67}{10}$$$$\frac{7x+4}{(x+1)(x^2+9)}=\frac{-3/10}{x+1}+\frac{\left(3/10\right)x+\left(67/10\right)}{x^2+9}$$$$=-\frac{3}{10(x+1)}+\frac{3x+67}{10(x^2+9)}$$$$$$

Commented by Rasheed.Sindhi last updated on 18/May/18

$$\mathrm{\top }\nparallel \mathrm{\forall }\cap \mid ⟨\mathrm{\$}!$$

Commented by abdo mathsup 649 cc last updated on 17/May/18

$$sirResheedthismethodisgeneralandeasy$$$$forawehaveF\left(x\right)=\frac{a}{x+1}+\frac{bx+c}{x^2+9}\Rightarrow$$$${lim}_{x\to -1}(x+1)F\left(x\right)=a+{lim}_{x\to -1}(x+1)\frac{bx+c}{x^2+9}$$$$=afromanotherside$$$${lim}_{x\to -1}(x+1)F\left(x\right)={lim}_{x\to -1}\frac{7x+4}{x^2+9}=\frac{-3}{10}....$$$$wedothesameforbandc...$$

Commented by abdo mathsup 649 cc last updated on 17/May/18

$$yesyesallthisprofilsareformebutnotallID$$$$containingabdo....$$

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