Evaluate the following integrals: x^2+1 x^2-5x+6 dx

Question

Evaluate the following integrals:

$\int\frac{\text{x}^2+1}{\text{x}^2-5\text{x}+6}\text{ dx}$

✓

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)

$\text{I}=\int\text{dx}-5\int\frac{\text{dx}}{\text{x}-2}+10\int\frac{\text{dx}}{\text{x}-3}$

$=\text{x}-5\log|\text{x}-2|+10\log|\text{x}-3|+\text{C}$

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