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2x+3/(x+3)(x^2+4) |
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Answer» \(\frac{2x+3}{(x+3)(x^2+4)}=\frac{A}{x+3}+\frac{Bx+C}{x^2+4}\) ⇒ 2x + 3 = A(x2+4) + (Bx+C) (x+3) ⇒ (A+B)x2 + (3B+C)x + 4A+3C = 2x+3 By comparing corresponding components, we get A+B = 0 ...(1) 3B+C = 2 ...(2) 4A+3C = 3 ...(3) Put B = -A from equation (1) into equation (2), we get -3A+C = 2 ⇒ -9A+3C = 6 ...(4) Subtract equation (4) from equation (3), we get 13A = -3 ⇒ A = -3/13 ∴ B = -A = 3/13 and C = 2 - 3B = 2 - 9/13 = 17/13 ∴ \(\frac{2x+3}{(x+3)(x^2+4)}=\frac{-3/13}{x+3}+\cfrac{\frac{3}{13}x+\frac{17}{13}}{x^2+4}\) |
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