Tiêu đề 4 Partial Fractions.pdf ✅

MSTC 4: PARTIAL FRACTIONS 1. Common Cases and their Decompositions CASE PARTIAL FRACTION ax + b (x − m)(x − n) = A x − m + B x − n ax + b (x − m) 3 = A x − m + B (x − m) 2 + C (x − m) 3 ax 2 + bx + c (x − l)(x − m)(x − n) = A x − l + B x − m + C x − n ax 2 + bx + c (x − l)(px 2 + qx + r) = A x − l + Bx + C px 2 + qx + r ax 2 + bx + c (x − m) 2(x − n) = A x − m + B (x − m) 2 + C x − n ax 3 + bx 2 + cx + d (px 2 + qx + r) 2 = Ax + B px 2 + qx + r + Cx + D (px 2 + qx + r) 2 - Example: Decompose the partial fraction 4x 3 − x 2 − 19 (x − 1)(x − 3)(x 2 + 2x + 5) = A x − 1 + B x − 3 + Cx + D x 2 + 2x + 5 [SOLUTION] 4x 3 − x 2 − 19 (x − 1)(x − 3)(x 2 + 2x + 5) = A x − 1 + B x − 3 + Cx + D x 2 + 2x + 5 4x 3 − x 2 − 19 (x − 1)(x − 3)(x 2 + 2x + 5) = A(x − 3)(x 2 + 2x + 5) + B(x − 1)(x 2 + 2x + 5) + (Cx + D)(x − 1)(x − 3) (x − 1)(x − 3)(x 2 + 2x + 5) 4x 3 − x 2 − 19 = A(x − 3)(x 2 + 2x + 5) + B(x − 1)(x 2 + 2x + 5) + (Cx + D)(x − 1)(x − 3) Set x = 1 (By setting x − 1 = 0) 4(1) 3 − (1) 2 − 19 = A(1 − 3)[(1) 2 + 2(1) + 5] −16 = −16A A = 1 Set x = 3 (By setting x − 3 = 0) 4(3) 3 − (3) 2 − 19 = B(3 − 1)[(3) 2 + 2(3) + 5] 80 = 40B B = 2 Set x = 0 (Random value) −19 = A(−3)(5) + B(−1)(5) + D(−1)(−3) −19 = (1)(−15) + 2(−5) + D(3) D = 2