Question
Evaluate the following integrals:

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

Answer

Let $\text{I}=\int\frac{2\text{x}+5}{\text{x}^2-\text{x}-2}\text{ dx}$
Let $2\text{x}+5=\lambda\frac{\text{d}}{\text{dx}}\big(\text{x}^2-\text{x}-2\big)+\mu$
$=\lambda(2\text{x}-1)+\mu$
$2\text{x}+5=(2\lambda)\text{x}-\lambda+\mu$
Comparing the coefficients of like power of x,
$2\lambda=2\Rightarrow\lambda=1$
$-\lambda+\mu=5\Rightarrow-1+\mu=5$
$\mu=6$
So, $\text{I}=\int\frac{(2\text{x}-1)+6}{\text{x}^2-\text{x}-2}\text{ dx}$
$\text{I}=\int\frac{(2\text{x}-1)}{\text{x}^2-\text{x}-2}\text{ dx}+6\int\frac{1}{\text{x}^2-2\text{x}\big(\frac{1}{2}\big)+\big(\frac{1}{2}\big)^2-\big(\frac{1}{2}\big)^2-2}\text{ dx}$
$\text{I}=\int\frac{2\text{x}-1}{\text{x}^2-\text{x}-2}\text{ dx}+6\int\frac{1}{\big(\text{x}-\frac{1}{2}\big)^2-\frac{9}{4}}\text{ dx}$
$\text{I}=\int\frac{2\text{x}-1}{\text{x}^2-\text{x}-2}\text{ dx}+6\int\frac{1}{\big(\text{x}-\frac{1}{2}\big)^2-\big(\frac{3}{2}\big)^2}\text{ dx}$
$\text{I}=\log\big|\text{x}^2-\text{x}-2\big|+\frac{6}{2\big(\frac{3}{2}\big)}\log\Bigg|\frac{\text{x}-\frac{1}{2}-\frac{3}{2}}{\text{x}-\frac{1}{2}+\frac{3}{2}}\Bigg|+\text{C}$ $\Big[\text{Since }\int\frac{1}{\text{x}^2+\text{a}^2}\text{ dx}=\frac{1}{2\text{a}}\log\Big|\frac{\text{x}-\text{a}}{\text{x}+\text{a}}\Big|+\text{C}\Big]$
$\text{I}=\log\big|\text{x}^2-\text{x}-2\big|+2\log\Big|\frac{\text{x}-2}{\text{x}+1}\Big|+\text{C}$

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