Evaluate the following intregals: 2x+1 (x+2)(x-3)^2 dx - Vidyadip

Question

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

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Answer

Let $\frac{2\text{x}+1}{(\text{x}+2)(\text{x}-3)^2}=\frac{\text{A}}{\text{x}+2}+\frac{\text{B}}{\text{x}-3}+\frac{\text{C}}{(\text{x}-3)^2}$
$\Rightarrow 2x + 1 = A (x - 3)^2 + B (x + 2) (x - 3) + C (x + 2)$
$= (A + B)x^2 + (-6A - B + C)x + (9A - 6B + 2C)$
Equating similar terms, we get, $A + B = 0 $
$\Rightarrow A = -B -6A - B +C = 2 $
$\Rightarrow 5B + C = 2 $
$9A - 6B + 2C = 1 $
$\Rightarrow -15B + 2C = 1$
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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