$\int \frac{a x^{2} + b x + c}{(x - a) (x - b) (x - c)} d x, where a, b, c are distinct$

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Solution

$We have, I = \int \frac{a x^{2} + b x + c}{(x - a) (x - b) (x - c)} d x Let \frac{a x^{2} + b x + c}{(x - a) (x - b) (x - c)} = \frac{A}{x - a} + \frac{B}{x - b} + \frac{C}{x - c} \Rightarrow a x^{2} + b x + c = A (x - b) (x - c) + B (x - c) (x - a) + C (x - a) (x - b) \Rightarrow a x^{2} + b x + c = A [x^{2} - (b + c) x + b c] + B [x^{2} - (c + a) x + c a] + C [x^{2} - (a + b) x + a b] \Rightarrow a x^{2} + b x + c = (A + B + C) x^{2} - [A (b + c) + B (c + a) + C (a + b)] x + A b c + B c a + C a b Equating the coefficients on both sides, we get a = A + B + C . . . . . (1) b = - [A (b + c) + B (c + a) + C (a + b)] . . . . . (2) c = A b c + B c a + C a b . . . . . (3) Solving (1), (2) and (3), we get A = \frac{a^{3} + a b + c}{(a - b) (a - c)} B = \frac{a b^{2} + b^{2} + c}{(b - a) (b - c)} C = \frac{a c^{2} + b c + c}{(c - a) (c - b)} ∴ I = \int [\frac{a^{3} + a b + c}{(a - b) (a - c)} \times \frac{1}{x - a} + \frac{a b^{2} + b^{2} + c}{(b - a) (b - c)} \times \frac{1}{x - b} + \frac{a c^{2} + b c + c}{(c - a) (c - b)} \times \frac{1}{x - c}] d x = \frac{a^{3} + a b + c}{(a - b) (a - c)} \log |x - a| + \frac{a b^{2} + b^{2} + c}{(b - a) (b - c)} \log |x - b| + \frac{a c^{2} + b c + c}{(c - a) (c - b)} \log |x - c| + K$

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