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Question Number 60320 by Sardor2211 last updated on 19/May/19

Commented by maxmathsup by imad last updated on 20/May/19

$$letI=\int \frac{x^3-5x^2+5x+23}{(x^2-1)(x-5)}dxletdecomposeF\left(x\right)=\frac{x^3-5x^2+5x+23}{(x^2-1)(x-5)}$$$$F\left(x\right)=\frac{x^3-5x^2+5x+23}{(x-1)(x+1)(x-5)}=1+\frac{a}{x-1}+\frac{b}{x+1}+\frac{c}{x-5}$$$$a={lim}_{x\to 1}(x-1)F\left(x\right)=\frac{1-5+5+23}{2(-4)}=\frac{24}{-8}=-3$$$$b={lim}_{x\to -1}(x+1)F\left(x\right)=\frac{-1-5-5+23}{(-2)(-6)}=\frac{12}{12}=1$$$$c={lim}_{x\to 5}(x-5)F\left(x\right)=\frac{5^3-5^3+25+23}{4.6}=\frac{48}{24}=2\Rightarrow$$$$F\left(x\right)=1-\frac{3}{x-1}+\frac{1}{x+1}+\frac{2}{x-5}\Rightarrow I=\int F\left(x\right)dx\Rightarrow$$$$I=x-3ln\mid x-1\mid +ln\mid x+1\mid +2ln\mid x-5\mid +C.$$$$$$

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