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Properties of Inequalities

∫ x 2+1 x x 2...

Question

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

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Solution

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

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