Question
Evaluate the following integrals:$\int\frac{\text{x}^2+1}{\text{x}^2-5\text{x}+6}\text{ dx}$
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Answer
Let $\text{I}=\int\Big(\frac{\text{x}^2+1}{\text{x}^2-5\text{x}+6}\Big)\text{dx}$ Dividing Numerator by Denominator
$\frac{\text{x}^2+1}{\text{x}^2-5\text{x}+6}=1+\Big(\frac{5\text{x}-5}{\text{x}^2-5\text{x}+6}\Big)\ ....(1)$ Also $\frac{5\text{x}-5}{\text{x}^2-5\text{x}+6}=\frac{5\text{x}-5}{(\text{x}-2)(\text{x}-3)}$ Let $\Rightarrow\frac{5\text{x}-5}{(\text{x}-2)(\text{x}-3)}=\frac{\text{A}}{\text{x}-2}+\frac{\text{B}}{\text{x}-3}$ $\Rightarrow\frac{5\text{x}-5}{(\text{x}-2)(\text{x}-3)}=\frac{\text{A}(\text{x}-3)+\text{B}(\text{x}-2)}{(\text{x}-2)(\text{x}-3)}$ $\Rightarrow5\text{x}-5=\text{A}(\text{x}-3)+\text{B}(\text{x}-2)$ Let $\text{x}=3$ $5\times3-5=\text{}A\times0+\text{B}(3-2)$ $10=\text{B}$ Let $\text{x}=2$ $5\times2-5=\text{}A(2-3)+\text{B}\times0$ $\text{A}=-5$ $\Rightarrow\frac{5\text{x}-5}{(\text{x}-2)(\text{x}-3)}=\frac{-5}{\text{x}-2}+\frac{10}{\text{x}-3}\ ... ..(2)$ From (1) and (2) $\text{I}=\int\text{dx}-5\int\frac{\text{dx}}{\text{x}-2}+10\int\frac{\text{dx}}{\text{x}-3}$ $=\text{x}-5\log|\text{x}-2|+10\log|\text{x}-3|+\text{C}$
$\frac{\text{x}^2+1}{\text{x}^2-5\text{x}+6}=1+\Big(\frac{5\text{x}-5}{\text{x}^2-5\text{x}+6}\Big)\ ....(1)$ Also $\frac{5\text{x}-5}{\text{x}^2-5\text{x}+6}=\frac{5\text{x}-5}{(\text{x}-2)(\text{x}-3)}$ Let $\Rightarrow\frac{5\text{x}-5}{(\text{x}-2)(\text{x}-3)}=\frac{\text{A}}{\text{x}-2}+\frac{\text{B}}{\text{x}-3}$ $\Rightarrow\frac{5\text{x}-5}{(\text{x}-2)(\text{x}-3)}=\frac{\text{A}(\text{x}-3)+\text{B}(\text{x}-2)}{(\text{x}-2)(\text{x}-3)}$ $\Rightarrow5\text{x}-5=\text{A}(\text{x}-3)+\text{B}(\text{x}-2)$ Let $\text{x}=3$ $5\times3-5=\text{}A\times0+\text{B}(3-2)$ $10=\text{B}$ Let $\text{x}=2$ $5\times2-5=\text{}A(2-3)+\text{B}\times0$ $\text{A}=-5$ $\Rightarrow\frac{5\text{x}-5}{(\text{x}-2)(\text{x}-3)}=\frac{-5}{\text{x}-2}+\frac{10}{\text{x}-3}\ ... ..(2)$ From (1) and (2) $\text{I}=\int\text{dx}-5\int\frac{\text{dx}}{\text{x}-2}+10\int\frac{\text{dx}}{\text{x}-3}$ $=\text{x}-5\log|\text{x}-2|+10\log|\text{x}-3|+\text{C}$
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