Multiply by (x 2 + l)(x + 2), x2 1 (x 2 + 1)(x + 2) = A x + A3 A1 + 22 x+2 x +1 in Visual Studio .NET

Encode QR Code in Visual Studio .NET Multiply by (x 2 + l)(x + 2), x2 1 (x 2 + 1)(x + 2) = A x + A3 A1 + 22 x+2 x +1

[ j 2]

Comparing coef cients of x 0 (the constant terms), 1 = A1 + 2A3 Comparing coef cients of x 2 , 2 5 f (x) dx will now include, besides terms arising from any linear factors, at least one term of the form 1 = A1 + A2 or A2 = 1 A1 = Ax + B dx = x 2 + bx + c A x+ b 2 +C AB 2 2 2 c b >0 4 c B or 1 1 8 A3 = (1 + A1 ) = 2 2 5 = 4 5

dx b 2 2 x+ + 2 b Au + C = du let u x + 2 u2 + 2 du u du +C =A u2 + 2 u2 + 2 A C 1 u = ln (u2 + 2 ) + tan 2

(For a guarantee that is a real number, see Problem 40.7.)

Case 4:

A repeated quadratic factor (x 2 + bx + c)k contributes to the partial fractions representation the expression A1 x + A2 A3 x + A 4 A2k 1 x + A2k + + + 2 + bx + c 2 + bx + c)2 ax (ax (ax 2 + bx + c)k The computations in this case may be long and tedious. EXAMPLE

Multiply by (x 2 + 1)2 , Compare coef cients of x 3 , 1 = A1 Compare coef cients of x 2 , 0 = A2 Compare coef cients of x, 0 = A1 + A3 Compare coef cients of x 0 , 1 = A2 + A4 The new contribution to Ax + B dx = A (x 2 + bx + c) j or A4 = 1 A 2 = 1 or A3 = A1 = 1 A x + A2 x3 + 1 A x + A4 = 12 + 32 (x 2 + 1)2 x +1 (x + 1)2 x 3 + 1 = (A1 x + A2 )(x 2 + 1) + A3 x + A4 (4)

f (x) dx will consist of one or more terms of the form u du du +C (u2 + 2 ) j (u2 + 2 ) j E + F cos2( j 1) d = 2 (u + 2 ) j 1 [as in Case 3] [let u = tan ]

and we know how to evaluate the trigonometric integral [see Problem 38.12(a) or example (b) of Section 39.1].

CHAP. 40]

The numerator has greater degree than the denominator. Therefore, divide the numerator by the denominator,

Thus, 2x 3 + x 2 6x + 7 7x + 1 = 2x 1 + 2 x2 + x 6 x +x 6 7x + 1 A1 A2 = + (x + 3)(x 2) x+3 x 2 Multiply by the denominator (x + 3)(x 2), 7x + 1 = A1 (x 2) + A2 (x + 3) Let x = 2, Let x = 3, Thus, 15 = 0 + 5A2 or A2 = 3 or A1 = 4 20 = 5A1 + 0 4 3 7x + 1 + = (x + 3)(x 2) x+3 x 2 2x 3 + x 2 6x + 7 dx = x2 + x 6 (2x 1)dx + 4 dx + x+3 3 dx x 2

Next, factor the denominator, x 2 + x 6 = (x + 3)(x 2). The partial fractions decomposition has the form (Case 1)

x 3 3x 2 9x + 27 = (x 3)(x 2 9) = (x 3)(x 3)(x + 3) = (x 3)2 (x + 3)