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Properties of Inequalities

∫ 2 x-3 x 2-1...

Question

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

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Solution

$\int \frac{2 x - 3}{(x^{2} - 1) (2 x + 3)} d x = \int \frac{(2 x - 3)}{(x - 1) (x + 1) (2 x + 3)} d x Let \frac{2 x - 3}{(x - 1) (x + 1) (2 x + 3)} = \frac{A}{x - 1} + \frac{B}{x + 1} + \frac{C}{2 x + 3} \Rightarrow \frac{2 x - 3}{(x - 1) (x + 1) (2 x + 3)} = \frac{A (x + 1) (2 x + 3) + B (x + 1) (2 x + 3) + C (x^{2} - 1)}{(x - 1) (x + 1) (2 x + 3)} \Rightarrow 2 x - 3 = A (x + 1) (2 x + 3) + B (x - 1) (2 x + 3) + C (x + 1) (x - 1) . . . (1) Putting x + 1 = 0 or x = - 1 in eq (1) \Rightarrow - 2 - 3 = B (- 1 - 1) (- 2 + 3) \Rightarrow - 5 = B (- 2) (1) \Rightarrow B = \frac{5}{2} Putting x - 1 = 0 or x = 1 in eq (1) \Rightarrow 2 - 3 = A (1 + 1) (2 + 3) \Rightarrow - 1 = A (2) (5) \Rightarrow A = \frac{- 1}{10} Putting 2 x + 3 = 0 or x = \frac{- 3}{2} in eq (1) \Rightarrow 2 \times - \frac{3}{2} - 3 = A \times 0 + B \times 0 + C (- \frac{3}{2} + 1) (\frac{- 3}{2} - 1) \Rightarrow - 6 = C (- \frac{1}{2}) (\frac{- 5}{2}) \Rightarrow C = - \frac{24}{5} ∴ \frac{2 x - 3}{(x - 1) (x + 1) (2 x + 3)} = \frac{- 1}{10 (x - 1)} + \frac{5}{2 (x + 1)} - \frac{24}{5 (2 x + 3)} \Rightarrow \int \frac{(2 x - 3)}{(x - 1) (x + 1) (2 x + 3)} d x = \frac{- 1}{10} \int \frac{1}{x - 1} d x + \frac{5}{2} \int \frac{1}{x + 1} d x - \frac{24}{5} \int \frac{1}{2 x + 3} d x = \frac{- 1}{10} \ln |x - 1| + \frac{5}{2} \ln |x + 1| - \frac{24}{5} \ln \frac{|2 x + 3|}{3} + C = - \frac{1}{10} \ln |x - 1| + \frac{5}{2} \ln |x + 1| - \frac{12}{5} \ln |2 x + 3| + C$

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