Question
Integrate the rational function in exercise:
$\frac{3\text{x}-1}{(\text{x}+2)^2}$

Answer

Let,  $\text{I}=\int\frac{3\text{x}-1}{(\text{x}+2)^2}\text{dx}\dots(\text{i})$Putting x + 2 = t
⇒ x = t - 2
$\frac{\text{dx}}{\text{dt}}=1$
⇒ dx = dt
Putting this value in eq. (i),
$\text{I}=\int\frac{3(\text{t}-2)-1}{(\text{t})^2}\text{dt}=\int\frac{3\text{t}-6-1}{\text{t}^2}\text{dt}=\int\frac{3\text{t}-7}{\text{t}^2}\text{dt}$
$=\int\Bigg(\frac{3\text{t}}{\text{t}^2}-\frac{7}{\text{t}^2}\Bigg)\text{dt}=\int\Bigg(\frac{3}{\text{t}}-\frac{7}{\text{t}^2}\Bigg)\text{dt}=3\int\frac{1}{\text{t}}\text{dt}-7\int\text{t}^{-2}\text{dt}$
$=3\text{log}|\text{t}|-7\frac{\text{t}^{-1}}{-1}+\text{c}=3\text{log}|\text{t}|+\frac{7}{\text{t}}+\text{c}=3\text{log}|\text{x}+2|+\frac{7}{\text{x}+2}+\text{c}$

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