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Question Number 60765 by readone97 last updated on 25/May/19

$$expressinpartialfraction14/(x^2+3)(x+2)$$

Commented by Prithwish sen last updated on 25/May/19

$$\frac{1}{({\mathrm{x}}^2+3)(\mathrm{x}+2)}=\frac{\mathrm{A}}{(\mathrm{x}+2)}+\frac{\mathrm{Bx}+\mathrm{C}}{({\mathrm{x}}^2+3)}$$

Commented by maxmathsup by imad last updated on 25/May/19

$$F\left(x\right)=\frac{14}{(x+2)(x^2+3)}=\frac{a}{x+2}+\frac{bx+c}{x^2+3}$$$$a={lim}_{x\to -2}(x+2)F\left(x\right)=\frac{14}{7}=2$$$${lim}_{x\to +\mathrm{\infty }}xF\left(x\right)=0=a+b\Rightarrow b=-a=-2\Rightarrow F\left(x\right)=\frac{2}{x+2}+\frac{-2x+c}{x^2+3}$$$$F\left(o\right)=\frac{14}{6}=\frac{7}{3}=1+\frac{c}{3}\Rightarrow 7=3+c\Rightarrow c=4\Rightarrow$$$$F\left(x\right)=\frac{2}{x+2}+\frac{-2x+4}{x^2+3}.$$

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