Question
Evaluate the following intregals:
$\int\frac{2\text{x}+1}{(\text{x}+2)(\text{x}-3)^2}\text{ dx}$

Answer

Let $\frac{2\text{x}+1}{(\text{x}+2)(\text{x}-3)^2}=\frac{\text{A}}{\text{x}+2}+\frac{\text{B}}{\text{x}-3}+\frac{\text{C}}{(\text{x}-3)^2}$

⇒ 2x + 1 = A (x - 3)2 + B (x + 2) (x - 3) + C (x + 2)

= (A + B)x2 + (-6A - B + C)x + (9A - 6B + 2C)

Equating similar terms, we get,

A + B = 0 ⇒ A = -B

-6A - B +C = 2 ⇒ 5B + C = 2

9A - 6B + 2C = 1 ⇒ -15B + 2C = 1

Solving, we get, $\text{B}=\frac{3}{25},\text{C}=\frac{7}{5},\text{A}=-\frac{3}{25}$

thus,

$\text{I}=-\frac{3}{25}\int\frac{\text{dx}}{\text{x}+2}+\frac{3}{25}\int\frac{\text{dx}}{\text{x}-3}+\frac{7}{5}\int\frac{\text{dx}}{(\text{x}-3)^2}$

$\text{I}=-\frac{3}{25}\log|\text{x}+2|+\frac{3}{25}\log|\text{x}-3|-\frac{7}{5(\text{x}-3)}+\text{C}$

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