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Special Integrals - 2

∫ 2 x+1 x-2 x...

Question

$\int \frac{2 x + 1}{(x - 2) (x - 3)} d x$

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Solution

$We have, I = \int \frac{(2 x + 1) d x}{(x - 2) (x - 3)} Let \frac{2 x + 1}{(x - 2) (x - 3)} = \frac{A}{x - 2} + \frac{B}{x - 3} \Rightarrow \frac{2 x + 1}{(x - 2) (x - 3)} = \frac{A (x - 3) + B (x - 2)}{(x - 2) (x - 3)} \Rightarrow 2 x + 1 = A (x - 3) + B (x - 2) Putting x - 3 = 0 \Rightarrow x = 3 ∴ 7 = A \times 0 + B \times (3 - 2) \Rightarrow B = 7 Putting x - 2 = 0 \Rightarrow x = 2 ∴ 5 = A (- 1) \Rightarrow A = - 5 ∴ I = - 5 \int \frac{d x}{x - 2} + 7 \int \frac{d x}{x - 3} = - 5 \log |x - 2| + 7 \log |x - 3| + C = \log {|x - 3|}^{7} - \log {|x - 2|}^{5} + C = \log |\frac{{(x - 3)}^{7}}{{(x - 2)}^{5}}| + C$

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