30.19
INTEGRATION OF RATIONAL FUNCTIONS: THE METHOD OF PARTIAL FRACTIONS
249
is irreducible
Then
Then
Then
1=9,4,
A=\.
Equate coefficients of x
3 : 1 = 2A + C + B.
Then D = -3 A -B-C=-W.
Thus,
In
Then
Equate coefficients ofx: 0 = A + B, B = — A — — \.
Equate coefficients of*: 0 = 3A + B + C+ D.
Equate coefficients of x:
0 = 2A + C + E,
E = 2A - C = - f.
(/)
(//)
Hence, the complete answer is
In
In
In
is a root of the denominator. When the latter is divided by x — \, we obtain x
2 + 3x + 2 =
Now,
and
cancels out, we are left with
Then 1 = A(x + 1) + B(x + 2). Let *=-!. Then 1 = B. Let x=-2. Then 1 = -A,
Thus,
In
30.20
Since x
2 + 5x + 6 = (x + 2)(x + 3), we have
Then x
2 + 2 = A(x +
Let x = -2. Then 6 = -2B,
Thus,
Let x = 0. Then 2 = 6,4, A =4.
Hence,
x
2 + x + l
(Z>
2 -4ac = -3<0).
*
3 + 1 = A(x
2 + x + I)
2 + x(x
2 + x + l)(Bx +C)'+ x(Dx + E).
Let
x = 0.
C= 1-2,4-5=1.
In
In
(oc + 2)(x + 1). Since x-\
x = \
= -In \x + 2\ + In |jc + l| + C, =
A=-l.-1.
2)(x + 3) + Bx(x + 3) + C*(* + 2).
S = -3.
Let jc=-3. Then 11 = 3C, C=^.
dx = § \n\x\ - 3 In |* + 2| + ^ In |;c + 3| + C,.