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Resolve the function \( f(x)=\frac{x^{3}}{\left(x^{2}+1\right)(x+1)} \) into partial fractions. |
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Answer» f(x) = \(\frac{x^3}{(x^2+1)(x+1)}\) = \(\frac{x^3}{x^3+x^2+x+1}\) = 1 - \(\frac{x^2+x+1}{(x^2+1)(x+1)}\) Now, \(\frac{x^2+x+1}{(x^2+1)(x+1)}\) = \(\frac{A}{x+1}\) + \(\frac{Bx+c}{x^2+1}\) ⇒ x2 + x + 1 = A(x2+1)+(Bx+c)(x+1) Put x = -1, we get 2A = 1-1+1 = 1 ⇒ A = \(\frac{1}{2}\) Put x = 0, we get A + C = 1 ⇒ C = 1 - A = 1 - \(\frac{1}{2}\) = \(\frac{1}{2}\) Put x = 1, we get 2A + 2B + 2C = 3 ⇒ 2C = 3 - 2A - 2B = 3 - 1 - 1 = 1 ⇒ C = \(\frac{1}{2}\) Hence, \(\frac{x^2+x+1}{(x^2+1)(x+1)}\) = \(\frac{1}{2}\)(\(\frac{1}{x+1}\) + \(\frac{x+1}{x^2+1}\)) ∴ f(x) = 1 - \(\frac{x^2+x+1}{(x^2+1)(x+1)}\) = 1 - \(\frac{1}{2}\)(\(\frac{1}{x+1}\) + \(\frac{x+1}{x^2+1}\)) |
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