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Higher Order Derivatives

∫ x 2 x-1 x+1...

Question

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

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Solution

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

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f (x, y) = ∣ ∣ ∣ ∣ ∣ \begin{matrix} y - x^{2} & x - y^{2} & x y - 1 x - y^{2} & x y - 1 & y - x^{2} x y - 1 & y - x^{2} & x - y^{2} \end{matrix} ∣ ∣ ∣ ∣ ∣

f (2, 2)

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