We #160;have,I= #8747;dxx4 1= #8747;dxx2 1 #160;x2+1= #8747;dxx 1 #160;x+1 #160;x2+1Let #160;1x 1 #160;x+1 #160;x2+1=Ax 1+Bx+1+Cx+Dx2+1 #8 ...

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Integration by Partial Fractions

∫ 1 × 4-1 dx

Question

$\int \frac{1}{x^{4} - 1} d x$

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Solution

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

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