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Special Integrals - 2

∫ x 2+12 x+1 ...

Question

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

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Solution

$We have, I = \int \frac{(x^{2} + 1) d x}{(2 x + 1) (x^{2} - 1)} = \int \frac{(x^{2} + 1) d x}{(2 x + 1) (x - 1) (x + 1)} Let \frac{(x^{2} + 1)}{(2 x + 1) (x - 1) (x + 1)} = \frac{A}{2 x + 1} + \frac{B}{x - 1} + \frac{C}{x + 1} \Rightarrow \frac{(x^{2} + 1)}{(2 x + 1) (x - 1) (x + 1)} = \frac{A (x^{2} - 1) + B (2 x + 1) (x + 1) + C (2 x + 1) (x - 1)}{(2 x + 1) (x - 1) (x + 1)} \Rightarrow x^{2} + 1 = A (x^{2} - 1) + B (2 x + 1) (x + 1) + C (2 x + 1) (x - 1) Putting x - 1 = 0 \Rightarrow x = 1 1 + 1 = A \times 0 + B (2 + 1) (1 + 1) + C \times 0 \Rightarrow 2 = B (3) (2) \Rightarrow B = \frac{1}{3} Putting x + 1 = 0 \Rightarrow x = - 1 1 + 1 = A \times 0 + B \times 0 + C (- 2 + 1) (- 1 - 1) \Rightarrow 2 = C (- 1) (- 2) \Rightarrow C = 1 Putting 2 x + 1 = 0 \Rightarrow x = - \frac{1}{2} {(- \frac{1}{2})}^{2} + 1 = A (\frac{1}{4} - 1) \Rightarrow \frac{1}{4} + 1 = A (- \frac{3}{4}) \Rightarrow \frac{5}{4} = A (- \frac{3}{4}) A = - \frac{5}{3} ∴ I = - \frac{5}{3} \int \frac{d x}{2 x + 1} + \frac{1}{3} \int \frac{d x}{x - 1} + \int \frac{d x}{x + 1} = - \frac{5}{3} \times \frac{\log |2 x + 1|}{2} + \frac{1}{3} \log |x - 1| + \log |x + 1| + C = - \frac{5}{6} \log |2 x + 1| + \frac{1}{3} \log |x - 1| + \log |x + 1| + C$

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