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Properties of Inequalities

∫ x 2+1 x 2-1...

Question

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

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Solution

$\int (\frac{x^{2} + 1}{x^{2} - 1}) d x = \int (\frac{x^{2} - 1 + 2}{x^{2} - 1}) d x = \int d x + 2 \int \frac{1}{x^{2} - 1^{2}} d x = \int d x + 2 \int \frac{1}{(x - 1) (x + 1)} d x . . . (1) ∴ \frac{1}{(x - 1) (x + 1)} = \frac{A}{x - 1} + \frac{B}{x + 1} \Rightarrow \frac{1}{(x - 1) (x + 1)} = \frac{A (x + 1) + B (x - 1)}{(x - 1) (x + 1)} \Rightarrow 1 = A (x + 1) B (x - 1) . . . . (2) Putting x + 1 = 0 or x = - 1 in eq (2) \Rightarrow 1 = A \times 0 + B (- 1 - 1) \Rightarrow B = \frac{- 1}{2} Putting x - 1 = 0 or x = 1 in eq (2) \Rightarrow 1 = A (1 + 1) + B \times 0 \Rightarrow A = \frac{1}{2} ∴ \frac{1}{(x - 1) (x + 1)} = \frac{1}{2 (x - 1)} - \frac{1}{2 (x + 1)} . . . (3) From eq (1) & eq (3) \int (\frac{x^{2} + 1}{x^{2} - 1}) d x = \int d x + 2 \int [\frac{1}{2 (x - 1)} - \frac{1}{2 (x + 1)}] d x = \int d x + \int \frac{d x}{x - 1} - \int \frac{d x}{x + 1} = x + \ln |x - 1| = - \ln |x + 1| + C = x + \ln |\frac{x - 1}{x + 1}| + C$

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