Question
Evaluate the following integrals:$\int\frac{\text{x}^2+\text{x}-1}{\text{x}^2+\text{x}-6}\text{ dx}$
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Answer
Let $\text{I}=\int\frac{\text{x}^2+\text{x}-1}{\text{x}^2+\text{x}-6}\text{ dx}$
$=\int\Big[1+\frac{5}{\text{x}^2+\text{x}-6}\Big]\text{dx}$
$=\text{x}+\int\frac{5}{\text{x}^2+\text{x}-6}\text{ dx}+\text{C}_1\ ....(1)$
$\text{I}_1=5\int\frac{1}{\text{x}^2+\text{x}-6}\text{ dx}$
$=5\int\frac{1}{\text{x}^2+2\text{x}\big(\frac{1}{2}\big)+\big(\frac{1}{2}\big)^2-\big(\frac{1}{2}\big)^2-6}\text{ dx}$
$=5\int\frac{1}{\big(\text{x}+\frac{1}{2}\big)^2-\big(\frac{5}{2}\big)^2}\text{ dx}$
$5\times\frac{1}{2\big(\frac{5}{2}\big)}\log\bigg|\frac{\text{x}+\frac{1}{2}-\frac{5}{2}}{\text{x}+\frac{1}{2}+\frac{5}{2}}\bigg|+\text{C}_2$
$\Big[\text{since},\int\frac{1}{\text{x}^2-\text{a}^2}\text{ dx}=\frac{1}{2\text{a}}\log\Big|\frac{\text{x}-\text{a}}{\text{x}+\text{a}}\Big|+\text{C}\Big]$
$\text{I}_1=\log\Big|\frac{\text{x}-2}{\text{x}+3}\Big|+\text{C}_2\ ....(2)$
Using equation (1) and (2)
$\text{I}=\text{x}+\log\Big|\frac{\text{x}-2}{\text{x}+3}\Big|+\text{C}$
$=\int\Big[1+\frac{5}{\text{x}^2+\text{x}-6}\Big]\text{dx}$
$=\text{x}+\int\frac{5}{\text{x}^2+\text{x}-6}\text{ dx}+\text{C}_1\ ....(1)$
$\text{I}_1=5\int\frac{1}{\text{x}^2+\text{x}-6}\text{ dx}$
$=5\int\frac{1}{\text{x}^2+2\text{x}\big(\frac{1}{2}\big)+\big(\frac{1}{2}\big)^2-\big(\frac{1}{2}\big)^2-6}\text{ dx}$
$=5\int\frac{1}{\big(\text{x}+\frac{1}{2}\big)^2-\big(\frac{5}{2}\big)^2}\text{ dx}$
$5\times\frac{1}{2\big(\frac{5}{2}\big)}\log\bigg|\frac{\text{x}+\frac{1}{2}-\frac{5}{2}}{\text{x}+\frac{1}{2}+\frac{5}{2}}\bigg|+\text{C}_2$
$\Big[\text{since},\int\frac{1}{\text{x}^2-\text{a}^2}\text{ dx}=\frac{1}{2\text{a}}\log\Big|\frac{\text{x}-\text{a}}{\text{x}+\text{a}}\Big|+\text{C}\Big]$
$\text{I}_1=\log\Big|\frac{\text{x}-2}{\text{x}+3}\Big|+\text{C}_2\ ....(2)$
Using equation (1) and (2)
$\text{I}=\text{x}+\log\Big|\frac{\text{x}-2}{\text{x}+3}\Big|+\text{C}$
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