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Harmonic Progression

∫ 3 x+5 x 3-x...

Question

$\int \frac{3 x + 5}{x^{3} - x^{2} - x + 1} d x$

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Solution

$We have, I = \int \frac{(3 x + 5) d x}{x^{3} - x^{2} - x + 1} = \int \frac{(3 x + 5) d x}{x^{2} (x - 1) - 1 (x - 1)} = \int \frac{(3 x + 5) d x}{(x^{2} - 1) (x - 1)} = \int \frac{(3 x + 5) d x}{(x - 1) (x + 1) (x - 1)} = \int \frac{(3 x + 5) d x}{{(x - 1)}^{2} (x + 1)} Let \frac{3 x + 5}{{(x - 1)}^{2} (x + 1)} = \frac{A}{x + 1} + \frac{B}{x - 1} + \frac{C}{{(x - 1)}^{2}} \Rightarrow \frac{3 x + 5}{{(x - 1)}^{2} (x + 1)} = \frac{A {(x - 1)}^{2} + B (x + 1) (x - 1) + C (x + 1)}{(x + 1) {(x - 1)}^{2}} \Rightarrow 3 x + 5 = A (x^{2} - 2 x + 1) + B (x^{2} - 1) + C x + C \Rightarrow 3 x + 5 = (A + B) x^{2} + (- 2 A + C) x + (A - B + C) Equating coefficient of like terms A + B = 0 . . . . . (1) - 2 A + C = 3 . . . . . (2) A - B + C = 5 . . . . . (3) Solving these three equations we get A = \frac{1}{2} B = - \frac{1}{2} C = 4 ∴ \frac{3 x + 5}{{(x - 1)}^{2} (x + 1)} = \frac{1}{2 (x + 1)} - \frac{1}{2 (x - 1)} + \frac{4}{{(x - 1)}^{2}} \Rightarrow I = \frac{1}{2} \int \frac{d x}{x + 1} - \frac{1}{2} \int \frac{d x}{x - 1} + 4 \int {(x - 1)}^{- 2} d x = \frac{1}{2} \log |x + 1| - \frac{1}{2} \log |x - 1| - \frac{4}{(x - 1)} + C' = \frac{1}{2} \log |\frac{x + 1}{x - 1}| - \frac{4}{x - 1} + C'$

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