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Question Number 108637 by mathdave last updated on 18/Aug/20

Answered by bemath last updated on 18/Aug/20

$$\frac{\mathrm{\angle }bemath\mathrm{\angle }}{⊸}$$$$\int dx+\int \frac{2dx}{(x-3)}+\int \frac{3dx}{x-2}=$$$$x+2\ln \mid x-3\mid +3\ln \mid x-2\mid +C$$

Answered by mathmax by abdo last updated on 18/Aug/20

$$\mathrm{I}=\int \frac{{\mathrm{x}}^2-7}{{\mathrm{x}}^2-5\mathrm{x}+6}\Rightarrow \mathrm{I}=\int \frac{{\mathrm{x}}^2-5\mathrm{x}+6+5\mathrm{x}-13}{{\mathrm{x}}^2-5\mathrm{x}+6}\mathrm{dx}$$$$=\mathrm{x}+\int \frac{5\mathrm{x}-13}{{\mathrm{x}}^2-5\mathrm{x}+6}\mathrm{dx}\mathrm{let}\mathrm{decompose}\mathrm{F}\left(\mathrm{x}\right)=\frac{5\mathrm{x}-13}{{\mathrm{x}}^2-5\mathrm{x}+6}$$$$\mathrm{\Delta }=25-24=1\Rightarrow {\mathrm{x}}_1=\frac{5+1}{2}=3\mathrm{and}{\mathrm{x}}_2=\frac{5-1}{2}=2\Rightarrow$$$$\mathrm{F}\left(\mathrm{x}\right)=\frac{5\mathrm{x}-13}{(\mathrm{x}-2)(\mathrm{x}-3)}=\frac{\mathrm{a}}{\mathrm{x}-2}+\frac{\mathrm{b}}{\mathrm{x}-3}$$$$\mathrm{a}=3\mathrm{and}\mathrm{b}=2\Rightarrow \mathrm{F}\left(\mathrm{x}\right)=\frac{3}{\mathrm{x}-2}+\frac{2}{\mathrm{x}-3}\Rightarrow$$$$\mathrm{I}=\mathrm{x}+\int (\frac{3}{\mathrm{x}-2}+\frac{2}{\mathrm{x}-3})\mathrm{dx}=\mathrm{x}+3\ln \mid \mathrm{x}-2\mid +2\ln \mid \mathrm{x}-3\mid +\mathrm{C}$$$$$$

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